# Calculus/Differentiation/Applications of Derivatives/Solutions
## Relative Extrema

Find the relative maximum(s) and minimum(s), if any, of the following functions.
$f(x)={\frac {x}{x+1}}\,$
$f'(x)={\frac {1}{(x+1)^{2}}}$
$f'(x)={\frac {1}{(x+1)^{2}}}$
$f(x)=(x-1)^{2/3}\,$
$f'(x)={\frac {2}{3}}(x-1)^{-1/3}={\frac {2}{3{\sqrt[{3}]{x-1}}}}$
$x=1$
$f(1)=0$
$f(x)$
$f'(x)={\frac {2}{3}}(x-1)^{-1/3}={\frac {2}{3{\sqrt[{3}]{x-1}}}}$
$x=1$
$f(1)=0$
$f(x)$
$f(x)=x^{2}+{\frac {2}{x}}\,$
$f'(x)=2x-{\frac {2}{x^{2}}}$
$2x-{\frac {2}{x^{2}}}=0\implies 2x^{3}-2=0\implies x=1$ $f^{\prime \prime }(x)=2+{\frac {4}{x^{3}}}$ $f^{\prime \prime }(1)=6$ Since the second derivative is positive, $x=1$ corresponds to a relative minimum.
$x=0$
$f'(x)=2x-{\frac {2}{x^{2}}}$
$2x-{\frac {2}{x^{2}}}=0\implies 2x^{3}-2=0\implies x=1$ $f^{\prime \prime }(x)=2+{\frac {4}{x^{3}}}$ $f^{\prime \prime }(1)=6$ Since the second derivative is positive, $x=1$ corresponds to a relative minimum.
$x=0$
$f(s)={\frac {s}{1+s^{2}}}\,$
$f'(x)={\frac {(1+s^{2})-2s^{2}}{(1+s^{2})^{2}}}={\frac {1-s^{2}}{(1+s^{2})^{2}}}$
${\frac {1-s^{2}}{(1+s^{2})^{2}}}=0\implies s=\pm 1$ $f^{\prime \prime }(x)={\frac {-2s(1+s^{2})^{2}-2(1+s^{2})(2s)(1-s^{2})}{(1+s^{2})^{4}}}$ $f^{\prime \prime }(-1)={\frac {-2(-1)(1+(-1)^{2})^{2}-2(1+(-1)^{2})(2(-1))(1-(-1)^{2})}{(1+(-1)^{2})^{4}}}={\frac {8}{16}}={\frac {1}{2}}$ Since the second derivative of $f(x)$ at $x=-1$ is positive, $x=-1$ corresponds to a relative mimimum. $f^{\prime \prime }(1)={\frac {-2(1+1^{2})^{2}-2(1+1^{2})(2)(1-1^{2})}{(1+1^{2})^{4}}}={\frac {-8}{16}}=-{\frac {1}{2}}$
$f(x)$
$x=1$
$f'(x)={\frac {(1+s^{2})-2s^{2}}{(1+s^{2})^{2}}}={\frac {1-s^{2}}{(1+s^{2})^{2}}}$
${\frac {1-s^{2}}{(1+s^{2})^{2}}}=0\implies s=\pm 1$ $f^{\prime \prime }(x)={\frac {-2s(1+s^{2})^{2}-2(1+s^{2})(2s)(1-s^{2})}{(1+s^{2})^{4}}}$ $f^{\prime \prime }(-1)={\frac {-2(-1)(1+(-1)^{2})^{2}-2(1+(-1)^{2})(2(-1))(1-(-1)^{2})}{(1+(-1)^{2})^{4}}}={\frac {8}{16}}={\frac {1}{2}}$ Since the second derivative of $f(x)$ at $x=-1$ is positive, $x=-1$ corresponds to a relative mimimum. $f^{\prime \prime }(1)={\frac {-2(1+1^{2})^{2}-2(1+1^{2})(2)(1-1^{2})}{(1+1^{2})^{4}}}={\frac {-8}{16}}=-{\frac {1}{2}}$
$f(x)$
$x=1$
$f(x)=x^{2}-4x+9\,$
$f'(x)=2x-4$
$2x-4=0\implies x=2$ $f^{\prime \prime }(x)=2$
$f'(x)=2x-4$
$2x-4=0\implies x=2$ $f^{\prime \prime }(x)=2$
$f(x)={\frac {x^{2}+x+1}{x^{2}-x+1}}\,$
${\begin{aligned}f'(x)&={\frac {(x^{2}-x+1)(2x+1)-(2x-1)(x^{2}+x+1)}{(x^{2}-x+1)^{2}}}\\&={\frac {2x^{3}-2x^{2}+2x+x^{2}-x+1-2x^{3}-2x^{2}-2x+x^{2}+x+1}{(x^{2}-x+1)^{2}}}\\&={\frac {-2x^{2}+2}{(x^{2}-x+1)^{2}}}\end{aligned}}$
${\frac {-2x^{2}+2}{(x^{2}-x+1)^{2}}}=0\implies x=\pm 1$ $f^{\prime \prime }(x)={\frac {(-4x)(x^{2}-x+1)^{2}-(2x-1)(-2x^{2}+2)}{(x^{2}-x+1)^{4}}}$ $f^{\prime \prime }(-1)={\frac {(-4(-1))((-1)^{2}-(-1)+1)^{2}-(2(-1)-1)(-2(-1)^{2}+2)}{((-1)^{2}-(-1)+1)^{4}}}={\frac {36}{81}}={\frac {4}{9}}$ Since $f^{\prime \prime }(-1)$ is positive, $x=-1$ corresponds to a relative minimum. $f^{\prime \prime }(1)={\frac {-4}{1}}=-4$
$f^{\prime \prime }(1)$
${\begin{aligned}f'(x)&={\frac {(x^{2}-x+1)(2x+1)-(2x-1)(x^{2}+x+1)}{(x^{2}-x+1)^{2}}}\\&={\frac {2x^{3}-2x^{2}+2x+x^{2}-x+1-2x^{3}-2x^{2}-2x+x^{2}+x+1}{(x^{2}-x+1)^{2}}}\\&={\frac {-2x^{2}+2}{(x^{2}-x+1)^{2}}}\end{aligned}}$
${\frac {-2x^{2}+2}{(x^{2}-x+1)^{2}}}=0\implies x=\pm 1$ $f^{\prime \prime }(x)={\frac {(-4x)(x^{2}-x+1)^{2}-(2x-1)(-2x^{2}+2)}{(x^{2}-x+1)^{4}}}$ $f^{\prime \prime }(-1)={\frac {(-4(-1))((-1)^{2}-(-1)+1)^{2}-(2(-1)-1)(-2(-1)^{2}+2)}{((-1)^{2}-(-1)+1)^{4}}}={\frac {36}{81}}={\frac {4}{9}}$ Since $f^{\prime \prime }(-1)$ is positive, $x=-1$ corresponds to a relative minimum. $f^{\prime \prime }(1)={\frac {-4}{1}}=-4$
$f^{\prime \prime }(1)$

## Range of Function

$x+1/x$
$f(x)=x+{\frac {1}{x}}$
$f'(x)=1-{\frac {1}{x^{2}}}$ $1-{\frac {1}{x^{2}}}=0\implies x=\pm 1$ $f^{\prime \prime }(x)={\frac {2}{x^{3}}}$ $f^{\prime \prime }(-1)=-2$ Since $f^{\prime \prime }(-1)$ is negative, $x=-1$ corresponds to a relative maximum. $f(-1)=-2$ $\lim \limits _{x\to -\infty }f(x)=-\infty$ For $x<-1$, $f'(x)$ is positive, which means that the function is increasing. Coming from very negative $x$-values, $f$ increases from a very negative value to reach a relative maximum of $-2$ at $x=-1$. For $-1<x<1$, $f'(x)$ is negative, which means that the function is decreasing. $\lim _{x\to 0^{-}}f(x)=-\infty$ $\lim _{x\to 0^{+}}f(x)=+\infty$ $f^{\prime \prime }(1)=2$ Since $f^{\prime \prime }(1)$ is positive, $x=1$ corresponds to a relative minimum. $f(1)=2$ Between $[-1,0)$ the function decreases from $-2$ to $-\infty$, then jumps to $+\infty$ and decreases until it reaches a relative minimum of $2$ at $x=1$. For $x>1$, $f'(x)$ is positive, so the function increases from a minimum of $2$.
$-2$
$2$
$f(x)=x+{\frac {1}{x}}$
$f'(x)=1-{\frac {1}{x^{2}}}$ $1-{\frac {1}{x^{2}}}=0\implies x=\pm 1$ $f^{\prime \prime }(x)={\frac {2}{x^{3}}}$ $f^{\prime \prime }(-1)=-2$ Since $f^{\prime \prime }(-1)$ is negative, $x=-1$ corresponds to a relative maximum. $f(-1)=-2$ $\lim \limits _{x\to -\infty }f(x)=-\infty$ For $x<-1$, $f'(x)$ is positive, which means that the function is increasing. Coming from very negative $x$-values, $f$ increases from a very negative value to reach a relative maximum of $-2$ at $x=-1$. For $-1<x<1$, $f'(x)$ is negative, which means that the function is decreasing. $\lim _{x\to 0^{-}}f(x)=-\infty$ $\lim _{x\to 0^{+}}f(x)=+\infty$ $f^{\prime \prime }(1)=2$ Since $f^{\prime \prime }(1)$ is positive, $x=1$ corresponds to a relative minimum. $f(1)=2$ Between $[-1,0)$ the function decreases from $-2$ to $-\infty$, then jumps to $+\infty$ and decreases until it reaches a relative minimum of $2$ at $x=1$. For $x>1$, $f'(x)$ is positive, so the function increases from a minimum of $2$.
$-2$
$2$

## Absolute Extrema

Determine the absolute maximum and minimum of the following functions on the given domain
$f(x)={\frac {1}{3}}x^{3}-{\frac {1}{2}}x^{2}+1$
$[0,3]$
$f$
$[0,3]$
$[0,3]$
$f'(x)=x^{2}-x$ $x^{2}-x=0\implies x=0,1$ $f(0)=1$ $f(1)={\frac {5}{6}}$ Check the endpoint: $f(3)={\frac {11}{2}}$
$f$
$[0,3]$
$[0,3]$
$f'(x)=x^{2}-x$ $x^{2}-x=0\implies x=0,1$ $f(0)=1$ $f(1)={\frac {5}{6}}$ Check the endpoint: $f(3)={\frac {11}{2}}$
$f(x)=({\frac {4}{3}}x^{2}-1)x$
$[-{\frac {1}{2}},2]$
$f'(x)=({\frac {4}{3}}x^{2}-1)+x({\frac {8}{3}}x)=4x^{2}-1$
$4x^{2}-1=0\implies x=\pm {\frac {1}{2}}$ $f(-{\frac {1}{2}})=({\frac {1}{3}}-1)(-{\frac {1}{2}})={\frac {1}{3}}$ $f({\frac {1}{2}})=({\frac {1}{3}}-1)({\frac {1}{2}})=-{\frac {1}{3}}$ $f(2)=({\frac {16}{3}}-1)(2)={\frac {26}{3}}$
$f'(x)=({\frac {4}{3}}x^{2}-1)+x({\frac {8}{3}}x)=4x^{2}-1$
$4x^{2}-1=0\implies x=\pm {\frac {1}{2}}$ $f(-{\frac {1}{2}})=({\frac {1}{3}}-1)(-{\frac {1}{2}})={\frac {1}{3}}$ $f({\frac {1}{2}})=({\frac {1}{3}}-1)({\frac {1}{2}})=-{\frac {1}{3}}$ $f(2)=({\frac {16}{3}}-1)(2)={\frac {26}{3}}$

## Determine Intervals of Change

Find the intervals where the following functions are increasing or decreasing
$f(x)=10-6x-2x^{2}$
$f'(x)=-6-4x$
$-6-4x=0\implies x=-{\frac {3}{2}}$ $f'(x)$ is the equation of a line with negative slope, so $f'(x)$ is positive for $x<-{\frac {3}{2}}$ and negative for $x>{\frac {3}{2}}$.
$f'(x)=-6-4x$
$-6-4x=0\implies x=-{\frac {3}{2}}$ $f'(x)$ is the equation of a line with negative slope, so $f'(x)$ is positive for $x<-{\frac {3}{2}}$ and negative for $x>{\frac {3}{2}}$.
$f(x)=2x^{3}-12x^{2}+18x+15$
$f'(x)=6x^{2}-24x+18$
${\begin{aligned}6x^{2}-24x+18=0&\implies x-4x+3=0\\&\implies (x-1)(x-3)=0\\&\implies x=1,3\end{aligned}}$ $f'(x)$ is the equation of a bowl-shaped parabola that crosses the $x$-axis at $1$ and $3$, so $f'(x)$ is negative for $1<x<3$ and positive elsewhere.
$f'(x)=6x^{2}-24x+18$
${\begin{aligned}6x^{2}-24x+18=0&\implies x-4x+3=0\\&\implies (x-1)(x-3)=0\\&\implies x=1,3\end{aligned}}$ $f'(x)$ is the equation of a bowl-shaped parabola that crosses the $x$-axis at $1$ and $3$, so $f'(x)$ is negative for $1<x<3$ and positive elsewhere.
$f(x)=5+36x+3x^{2}-2x^{3}$
$f'(x)=36+6x-6x^{2}$
${\begin{aligned}36+6x-6x^{2}=0&\implies 6+x-x^{2}=0\\&\implies (x+2)(-x+3)=0\\&\implies x=-2,3\end{aligned}}$ $f'(x)$ is the equation of a hill-shaped parabola that crosses the $x$-axis at $-2$ and $3$, so $f'(x)$ is positive for $-2<x<3$ and negative elsewhere.
$f'(x)=36+6x-6x^{2}$
${\begin{aligned}36+6x-6x^{2}=0&\implies 6+x-x^{2}=0\\&\implies (x+2)(-x+3)=0\\&\implies x=-2,3\end{aligned}}$ $f'(x)$ is the equation of a hill-shaped parabola that crosses the $x$-axis at $-2$ and $3$, so $f'(x)$ is positive for $-2<x<3$ and negative elsewhere.
$f(x)=8+36x+3x^{2}-2x^{3}$
$f(x)=5x^{3}-15x^{2}-120x+3$
$f'(x)=15x^{2}-30x-120$
${\begin{aligned}15x^{2}-30x-120=0&\implies x^{2}-2x-8=0\\&\implies (x+2)(x-4)=0\\&\implies x=-2,4\end{aligned}}$ $f'$ is negative on $(-2,4)$ and positive elsewhere.
$f$
$f'(x)=15x^{2}-30x-120$
${\begin{aligned}15x^{2}-30x-120=0&\implies x^{2}-2x-8=0\\&\implies (x+2)(x-4)=0\\&\implies x=-2,4\end{aligned}}$ $f'$ is negative on $(-2,4)$ and positive elsewhere.
$f$
$f(x)=x^{3}-6x^{2}-36x+2$
$f'(x)=3x^{2}-12x-36$
${\begin{aligned}3x^{2}-12x-36=0&\implies x^{2}-4x-12=0\\&\implies (x+2)(x-6)=0\\&\implies x=-2,6\end{aligned}}$
$f$
$f'(x)=3x^{2}-12x-36$
${\begin{aligned}3x^{2}-12x-36=0&\implies x^{2}-4x-12=0\\&\implies (x+2)(x-6)=0\\&\implies x=-2,6\end{aligned}}$
$f$

## Determine Intervals of Concavity

Find the intervals where the following functions are concave up or concave down
$f(x)=10-6x-2x^{2}$
$f'(x)=-6-4x$
$f''(x)=-4$
$f'(x)=-6-4x$
$f''(x)=-4$
$f(x)=2x^{3}-12x^{2}+18x+15$
$f'(x)=6x^{2}-24x+18$
$f^{\prime \prime }(x)=12x-24$ $12x-24=0\implies x=2$ When $x<2$, $f^{\prime \prime }(x)$ is negative, and when $x>2$, $f^{\prime \prime }(x)$ is positive.
$f'(x)=6x^{2}-24x+18$
$f^{\prime \prime }(x)=12x-24$ $12x-24=0\implies x=2$ When $x<2$, $f^{\prime \prime }(x)$ is negative, and when $x>2$, $f^{\prime \prime }(x)$ is positive.
$f(x)=5+36x+3x^{2}-2x^{3}$
$f'(x)=36+6x-6x^{2}$
$f^{\prime \prime }(x)=6-12x$ $6-12x=0\implies x={\frac {1}{2}}$ $f^{\prime \prime }(x)$ is positive when $x<{\frac {1}{2}}$ and negative when $x>{\frac {1}{2}}$.
$f'(x)=36+6x-6x^{2}$
$f^{\prime \prime }(x)=6-12x$ $6-12x=0\implies x={\frac {1}{2}}$ $f^{\prime \prime }(x)$ is positive when $x<{\frac {1}{2}}$ and negative when $x>{\frac {1}{2}}$.
$f(x)=8+36x+3x^{2}-2x^{3}$
$f(x)=5x^{3}-15x^{2}-120x+3$
$f'(x)=15x^{2}-30x-120$
$f^{\prime \prime }(x)=30x-30$ $30x-30=0\implies x=1$ $f^{\prime \prime }(x)$ is positive when $x>1$ and negative when $x<1$.
$f'(x)=15x^{2}-30x-120$
$f^{\prime \prime }(x)=30x-30$ $30x-30=0\implies x=1$ $f^{\prime \prime }(x)$ is positive when $x>1$ and negative when $x<1$.
$f(x)=x^{3}-6x^{2}-36x+2$
$f'(x)=3x^{2}-12x-36$
$f^{\prime \prime }(x)=6x-12$ $6x-12=0\implies x=2$ $f^{\prime \prime }(x)$ is positive when $x>2$ and negative when $x<2$.
$f'(x)=3x^{2}-12x-36$
$f^{\prime \prime }(x)=6x-12$ $6x-12=0\implies x=2$ $f^{\prime \prime }(x)$ is positive when $x>2$ and negative when $x<2$.

## Word Problems

$4t$
$t$
${\frac {dx}{dt}}=v(t)=4t-6$ After 4 seconds, the rate of change in position with respect to time is
$v(4)=16-6=\mathbf {10{\frac {m}{s}}}$
${\frac {dx}{dt}}=v(t)=4t-6$ After 4 seconds, the rate of change in position with respect to time is
$v(4)=16-6=\mathbf {10{\frac {m}{s}}}$
$+y$
$y={\frac {12}{5}}x$ The distance between the bikes is given by $s={\sqrt {x^{2}+y^{2}}}={\sqrt {x^{2}+({\frac {12}{5}}x)^{2}}}={\sqrt {{\frac {144+25}{25}}x^{2}}}={\sqrt {{\frac {169}{25}}x^{2}}}={\frac {13}{5}}x$ Let $t$ represent the elapsed time in hours. We want ${\frac {ds}{dt}}$ when $t=1$. Apply the chain rule to $s$: ${\frac {ds}{dt}}={\frac {ds}{dx}}{\frac {dx}{dt}}={\frac {13}{5}}\cdot 5=13$
$+y$
$y={\frac {12}{5}}x$ The distance between the bikes is given by $s={\sqrt {x^{2}+y^{2}}}={\sqrt {x^{2}+({\frac {12}{5}}x)^{2}}}={\sqrt {{\frac {144+25}{25}}x^{2}}}={\sqrt {{\frac {169}{25}}x^{2}}}={\frac {13}{5}}x$ Let $t$ represent the elapsed time in hours. We want ${\frac {ds}{dt}}$ when $t=1$. Apply the chain rule to $s$: ${\frac {ds}{dt}}={\frac {ds}{dx}}{\frac {dx}{dt}}={\frac {13}{5}}\cdot 5=13$
$^{3}$
$^{2}$
$^{2}$
$r$
$h$
$V=\pi r^{2}h$ We are constricted to have a can with a volume of $200m^{3}$, so we use this fact to relate the radius and the height: $\pi r^{2}h=200\implies h={\frac {200}{\pi r^{2}}}$ The surface area of the side is $A_{side}=2\pi rh=2\pi r{\frac {200}{\pi r^{2}}}={\frac {400}{r}}$ and the cost of the side is $C_{side}={\frac {4000}{r}}$ The surface area of the top and bottom (which is also the cost) is $A_{tb}=C_{tb}=2\pi r^{2}$ The total cost is given by $C={\frac {4000}{r}}+2\pi r^{2}$ We want to minimize $C$, so take the derivative: $C'=-{\frac {4000}{r^{2}}}+4\pi r$ Find the critical points: $-{\frac {4000}{r^{2}}}+4\pi r=0\implies 4\pi r={\frac {4000}{r^{2}}}\implies 4\pi r^{3}=4000\implies r={\sqrt[{3}]{\frac {1000}{\pi }}}={\frac {10}{\sqrt[{3}]{\pi }}}$ Check the second derivative to see if this point corresponds to a maximum or minimum: $C^{\prime \prime }={\frac {8000}{r^{3}}}+4\pi$ $C^{\prime \prime }({\frac {10}{\sqrt[{3}]{\pi }}})={\frac {8000}{1000/\pi }}+4\pi =8\pi +4\pi =12\pi$ Since the second derivative is positive, the critical point corresponds to a minimum. Thus, the minimum cost is
$C({\frac {10}{\sqrt[{3}]{\pi }}})={\frac {4000}{10/{\sqrt[{3}]{\pi }}}}+2\pi {\frac {10}{\sqrt[{3}]{\pi }}}\approx \mathbf {\$878.76}$
$r$
$h$
$V=\pi r^{2}h$ We are constricted to have a can with a volume of $200m^{3}$, so we use this fact to relate the radius and the height: $\pi r^{2}h=200\implies h={\frac {200}{\pi r^{2}}}$ The surface area of the side is $A_{side}=2\pi rh=2\pi r{\frac {200}{\pi r^{2}}}={\frac {400}{r}}$ and the cost of the side is $C_{side}={\frac {4000}{r}}$ The surface area of the top and bottom (which is also the cost) is $A_{tb}=C_{tb}=2\pi r^{2}$ The total cost is given by $C={\frac {4000}{r}}+2\pi r^{2}$ We want to minimize $C$, so take the derivative: $C'=-{\frac {4000}{r^{2}}}+4\pi r$ Find the critical points: $-{\frac {4000}{r^{2}}}+4\pi r=0\implies 4\pi r={\frac {4000}{r^{2}}}\implies 4\pi r^{3}=4000\implies r={\sqrt[{3}]{\frac {1000}{\pi }}}={\frac {10}{\sqrt[{3}]{\pi }}}$ Check the second derivative to see if this point corresponds to a maximum or minimum: $C^{\prime \prime }={\frac {8000}{r^{3}}}+4\pi$ $C^{\prime \prime }({\frac {10}{\sqrt[{3}]{\pi }}})={\frac {8000}{1000/\pi }}+4\pi =8\pi +4\pi =12\pi$ Since the second derivative is positive, the critical point corresponds to a minimum. Thus, the minimum cost is
$C({\frac {10}{\sqrt[{3}]{\pi }}})={\frac {4000}{10/{\sqrt[{3}]{\pi }}}}+2\pi {\frac {10}{\sqrt[{3}]{\pi }}}\approx \mathbf {\$878.76}$
$216{\rm {\,cm}}$
$A=3lw$
$3w+4l=216$
Notice that $3w+4l=216\Leftrightarrow 3w=216-4l$, so $A=3lw=l(3w)\Rightarrow A(l)=l(216-4l)=216l-4l^{2}$. Therefore, to maximize the total area (which will also maximize the internal roaming areas for the animal since each area is congruent), we take the derivative of the area function with respect to length.
    $A^{\prime }(l)=216-8l$
    $A^{\prime }(l)=0\Rightarrow l=27$
Notice that since $A^{\prime \prime }(l)=-8<0$ for all $l$, it implies that $A(27)$ is a local maximum.
    $A(27)=27(216-4\cdot 27)=2916$
$2916/3=\mathbf {972} {\rm {\,cm^{2}}}$
$A=3lw$
$3w+4l=216$
Notice that $3w+4l=216\Leftrightarrow 3w=216-4l$, so $A=3lw=l(3w)\Rightarrow A(l)=l(216-4l)=216l-4l^{2}$. Therefore, to maximize the total area (which will also maximize the internal roaming areas for the animal since each area is congruent), we take the derivative of the area function with respect to length.
    $A^{\prime }(l)=216-8l$
    $A^{\prime }(l)=0\Rightarrow l=27$
Notice that since $A^{\prime \prime }(l)=-8<0$ for all $l$, it implies that $A(27)$ is a local maximum.
    $A(27)=27(216-4\cdot 27)=2916$
$2916/3=\mathbf {972} {\rm {\,cm^{2}}}$
$r$
$(x,y)$
$x^{2}+y^{2}=r^{2}$
Since it must be the case that $x^{2}+y^{2}=r^{2}$, it also means that the set of points $P=\{(x,y),(-x,y),(x,-y),(-x,-y)\}$ also satisfy the relation. The distance between two horizontally or vertically related points is thus $2x$ and $2y$ respectively, so the area of the rectangle is given by $A=2x\cdot 2y=4xy$.
Therefore, in terms of $x$, the area of the rectangle is $A(x)=4x{\sqrt {r^{2}-x^{2}}}$ where $0\leq x\leq r$. Taking the derivative to find the maximum,
    $A^{\prime }(x)=4{\sqrt {r^{2}-x^{2}}}-{\frac {4x^{2}}{\sqrt {r^{2}-x^{2}}}}$
Notice that there exists a critical point on $x=\pm r$ (since the derivative does not exist on those points). We only worry about $x=r$ since $x=-r$ is outside of the points of interest.
Set the derivative to zero to find the other critical points.
    ${\begin{aligned}A^{\prime }(x)=0&\Rightarrow 4{\sqrt {r^{2}-x^{2}}}-{\frac {4x^{2}}{\sqrt {r^{2}-x^{2}}}}=0\\&\Rightarrow 4\left(r^{2}-x^{2}\right)-4x^{2}=0\\&\Rightarrow 4r^{2}-8x^{2}=0\\&\Rightarrow x^{2}={\frac {r^{2}}{2}}{\text{ or }}x=\pm {\frac {r}{\sqrt {2}}}\end{aligned}}$
Since $x=-{\frac {r}{\sqrt {2}}}$ falls outside our points of interest, the only critical points we care about is at $x={\frac {r}{\sqrt {2}}}$. Therefore, we test the three critical points:
    ${\begin{aligned}A(0)&=4\cdot (0)\cdot {\sqrt {r^{2}-(0)^{2}}}=0\\A\left({\frac {r}{\sqrt {2}}}\right)&=4\cdot \left({\frac {r}{\sqrt {2}}}\right)\cdot {\sqrt {r^{2}-\left({\frac {r}{\sqrt {2}}}\right)^{2}}}\\&=4\cdot {\frac {r}{\sqrt {2}}}\cdot {\sqrt {\frac {r^{2}}{2}}}\\&=4\cdot {\frac {r^{2}}{2}}=2r^{2}\\A(r)&=4\cdot (r)\cdot {\sqrt {r^{2}-(r)^{2}}}=0\end{aligned}}$
${\textbf {2r}}^{\textbf {2}}$
$(x,y)$
$x^{2}+y^{2}=r^{2}$
Since it must be the case that $x^{2}+y^{2}=r^{2}$, it also means that the set of points $P=\{(x,y),(-x,y),(x,-y),(-x,-y)\}$ also satisfy the relation. The distance between two horizontally or vertically related points is thus $2x$ and $2y$ respectively, so the area of the rectangle is given by $A=2x\cdot 2y=4xy$.
Therefore, in terms of $x$, the area of the rectangle is $A(x)=4x{\sqrt {r^{2}-x^{2}}}$ where $0\leq x\leq r$. Taking the derivative to find the maximum,
    $A^{\prime }(x)=4{\sqrt {r^{2}-x^{2}}}-{\frac {4x^{2}}{\sqrt {r^{2}-x^{2}}}}$
Notice that there exists a critical point on $x=\pm r$ (since the derivative does not exist on those points). We only worry about $x=r$ since $x=-r$ is outside of the points of interest.
Set the derivative to zero to find the other critical points.
    ${\begin{aligned}A^{\prime }(x)=0&\Rightarrow 4{\sqrt {r^{2}-x^{2}}}-{\frac {4x^{2}}{\sqrt {r^{2}-x^{2}}}}=0\\&\Rightarrow 4\left(r^{2}-x^{2}\right)-4x^{2}=0\\&\Rightarrow 4r^{2}-8x^{2}=0\\&\Rightarrow x^{2}={\frac {r^{2}}{2}}{\text{ or }}x=\pm {\frac {r}{\sqrt {2}}}\end{aligned}}$
Since $x=-{\frac {r}{\sqrt {2}}}$ falls outside our points of interest, the only critical points we care about is at $x={\frac {r}{\sqrt {2}}}$. Therefore, we test the three critical points:
    ${\begin{aligned}A(0)&=4\cdot (0)\cdot {\sqrt {r^{2}-(0)^{2}}}=0\\A\left({\frac {r}{\sqrt {2}}}\right)&=4\cdot \left({\frac {r}{\sqrt {2}}}\right)\cdot {\sqrt {r^{2}-\left({\frac {r}{\sqrt {2}}}\right)^{2}}}\\&=4\cdot {\frac {r}{\sqrt {2}}}\cdot {\sqrt {\frac {r^{2}}{2}}}\\&=4\cdot {\frac {r^{2}}{2}}=2r^{2}\\A(r)&=4\cdot (r)\cdot {\sqrt {r^{2}-(r)^{2}}}=0\end{aligned}}$
${\textbf {2r}}^{\textbf {2}}$
$R=3{\rm {\,in}}$
$O$
$h$
$r$
$r^{2}+\left({\frac {h}{2}}\right)^{2}=R^{2}$
Given the volume of a cylinder is $V=\pi r^{2}h$, and $r^{2}=R^{2}-\left({\frac {h}{2}}\right)^{2}=R^{2}-{\frac {h^{2}}{4}}$, the volume of the cylinder with respect to the height is
    $V(h)=\pi \left(R^{2}-{\frac {h^{2}}{4}}\right)h=\pi hR^{2}-{\frac {h^{3}\pi }{4}}$
    $V^{\prime }(h)=\pi R^{2}-{\frac {3h^{2}\pi }{4}}$
Setting the derivative to zero yields the critical points.
    ${\begin{aligned}V^{\prime }(h)=0&\Rightarrow \pi R^{2}-{\frac {3h^{2}\pi }{4}}=0\\&\Rightarrow \pi \left(R^{2}-{\frac {3h^{2}}{4}}\right)=0\\&\Rightarrow R^{2}-{\frac {3h^{2}}{4}}=0\Rightarrow R^{2}={\frac {3h^{2}}{4}}\\&\Rightarrow {\frac {4}{3}}R^{2}=h^{2}\\&\Rightarrow h=\pm {\sqrt {{\frac {4}{3}}R^{2}}}\end{aligned}}$.
Keep in mind that the volume must be positive, so $h>0$. As such, we only need to look at the critical point $h={\sqrt {{\frac {4}{3}}R^{2}}}$. Taking the second derivative
    $V^{\prime \prime }(h)=-{\frac {6\pi }{4}}h=-{\frac {3\pi }{2}}h$
Since the critical point $h={\sqrt {{\frac {4}{3}}R^{2}}}$ is positive, and $V^{\prime \prime }(h)$ is linear, $V^{\prime \prime }(0)=0$, and $V^{\prime \prime }(h)$ is trending downward,
    $V^{\prime \prime }\left({\sqrt {{\frac {4}{3}}R^{2}}}\right)<0$, so by the second derivative test, $V\left({\sqrt {{\frac {4}{3}}R^{2}}}\right)$ is a local maximum.
Recall the radius of the sphere is $R=3$, so $h={\sqrt {{\frac {4}{3}}\cdot 3^{2}}}={\sqrt {12}}$.
    ${\begin{aligned}V({\sqrt {12}})&=\pi \left(3^{2}-{\frac {\left({\sqrt {12}}\right)^{2}}{4}}\right){\sqrt {12}}\\&=\pi \left(9-3\right){\sqrt {4\cdot 3}}\\&=\pi \left(6\right)2{\sqrt {3}}=12\pi {\sqrt {3}}\end{aligned}}$
${\textbf {12}}{\boldsymbol {\pi }}{\sqrt {\textbf {3}}}\,{\textbf {in}}^{\textbf {3}}$
$O$
$h$
$r$
$r^{2}+\left({\frac {h}{2}}\right)^{2}=R^{2}$
Given the volume of a cylinder is $V=\pi r^{2}h$, and $r^{2}=R^{2}-\left({\frac {h}{2}}\right)^{2}=R^{2}-{\frac {h^{2}}{4}}$, the volume of the cylinder with respect to the height is
    $V(h)=\pi \left(R^{2}-{\frac {h^{2}}{4}}\right)h=\pi hR^{2}-{\frac {h^{3}\pi }{4}}$
    $V^{\prime }(h)=\pi R^{2}-{\frac {3h^{2}\pi }{4}}$
Setting the derivative to zero yields the critical points.
    ${\begin{aligned}V^{\prime }(h)=0&\Rightarrow \pi R^{2}-{\frac {3h^{2}\pi }{4}}=0\\&\Rightarrow \pi \left(R^{2}-{\frac {3h^{2}}{4}}\right)=0\\&\Rightarrow R^{2}-{\frac {3h^{2}}{4}}=0\Rightarrow R^{2}={\frac {3h^{2}}{4}}\\&\Rightarrow {\frac {4}{3}}R^{2}=h^{2}\\&\Rightarrow h=\pm {\sqrt {{\frac {4}{3}}R^{2}}}\end{aligned}}$.
Keep in mind that the volume must be positive, so $h>0$. As such, we only need to look at the critical point $h={\sqrt {{\frac {4}{3}}R^{2}}}$. Taking the second derivative
    $V^{\prime \prime }(h)=-{\frac {6\pi }{4}}h=-{\frac {3\pi }{2}}h$
Since the critical point $h={\sqrt {{\frac {4}{3}}R^{2}}}$ is positive, and $V^{\prime \prime }(h)$ is linear, $V^{\prime \prime }(0)=0$, and $V^{\prime \prime }(h)$ is trending downward,
    $V^{\prime \prime }\left({\sqrt {{\frac {4}{3}}R^{2}}}\right)<0$, so by the second derivative test, $V\left({\sqrt {{\frac {4}{3}}R^{2}}}\right)$ is a local maximum.
Recall the radius of the sphere is $R=3$, so $h={\sqrt {{\frac {4}{3}}\cdot 3^{2}}}={\sqrt {12}}$.
    ${\begin{aligned}V({\sqrt {12}})&=\pi \left(3^{2}-{\frac {\left({\sqrt {12}}\right)^{2}}{4}}\right){\sqrt {12}}\\&=\pi \left(9-3\right){\sqrt {4\cdot 3}}\\&=\pi \left(6\right)2{\sqrt {3}}=12\pi {\sqrt {3}}\end{aligned}}$
${\textbf {12}}{\boldsymbol {\pi }}{\sqrt {\textbf {3}}}\,{\textbf {in}}^{\textbf {3}}$
$6\,{\rm {ft}}$
$15$
$6$
The light and the man are related by a triangle relation as shown. The man, ${\overline {PB}}$, is moving away from the light at a rate of $6$ feet per second, so the length of ${\overline {AB}}$ must be changing. Therefore, set that length to be $x$. Assuming the length of the shadow and the distance from the post is $L$, i.e. $m{\overline {AQ}}=L$, the shadow has $m{\overline {BQ}}=L-x$.
Notice the similar triangles involved: $\triangle OCP\sim \triangle PBQ$ because $\angle OPC\cong \angle PQB$ (show this yourself!). As such, we may relate lengths as shown:
    ${\begin{aligned}{\frac {m{\overline {BQ}}}{m{\overline {PB}}}}&={\frac {m{\overline {CP}}}{m{\overline {OC}}}}\\{\frac {L-x}{6}}&={\frac {x}{15-6}}\\(L-x)(15-6)=9L-9x&=6x\\L&={\frac {15}{9}}x\end{aligned}}$
$S(x)=L-x={\frac {15}{9}}x-x={\frac {2}{3}}x$
${\frac {dS}{dt}}={\frac {2}{3}}{\frac {dx}{dt}}={\frac {2}{3}}\cdot 6\,{\rm {ft/s}}={\textbf {4}}\,{\textbf {ft/s}}$
The light and the man are related by a triangle relation as shown. The man, ${\overline {PB}}$, is moving away from the light at a rate of $6$ feet per second, so the length of ${\overline {AB}}$ must be changing. Therefore, set that length to be $x$. Assuming the length of the shadow and the distance from the post is $L$, i.e. $m{\overline {AQ}}=L$, the shadow has $m{\overline {BQ}}=L-x$.
Notice the similar triangles involved: $\triangle OCP\sim \triangle PBQ$ because $\angle OPC\cong \angle PQB$ (show this yourself!). As such, we may relate lengths as shown:
    ${\begin{aligned}{\frac {m{\overline {BQ}}}{m{\overline {PB}}}}&={\frac {m{\overline {CP}}}{m{\overline {OC}}}}\\{\frac {L-x}{6}}&={\frac {x}{15-6}}\\(L-x)(15-6)=9L-9x&=6x\\L&={\frac {15}{9}}x\end{aligned}}$
$S(x)=L-x={\frac {15}{9}}x-x={\frac {2}{3}}x$
${\frac {dS}{dt}}={\frac {2}{3}}{\frac {dx}{dt}}={\frac {2}{3}}\cdot 6\,{\rm {ft/s}}={\textbf {4}}\,{\textbf {ft/s}}$
29. A canoe is being pulled toward a dock (normal to the water) using a taut rope. The canoe is normal to the water while it is being pulled. The rope is hauled in at a constant $5\,{\rm {ft/s}}$. The dock is $5\,{\rm {ft}}$ above the water. Answer items (a) through (b).
$13\,{\rm {ft}}$
$x$
$y=5\,{\rm {ft}}$
$r=13\,{\rm {ft}}$
$x$
$y$
    $x^{2}+5^{2}=13^{2}\Rightarrow x=12\,{\rm {ft}}$.
Implicitly taking the derivative of the relation ($x^{2}+y^{2}=r^{2}$) tells us that $2x\cdot {\frac {dx}{dt}}=2r\cdot {\frac {dr}{dt}}$ (notice that the height of the dock is constant). Isolating for ${\frac {dx}{dt}}={\dot {x}}$:
    ${\begin{aligned}{\dot {x}}&={\frac {r}{x}}{\dot {r}}\\&={\frac {13}{5}}\cdot 5\,{\rm {ft/s}}\\&={\textbf {13}}\,{\textbf {ft/s}}\end{aligned}}$
$x$
$y=5\,{\rm {ft}}$
$r=13\,{\rm {ft}}$
$x$
$y$
    $x^{2}+5^{2}=13^{2}\Rightarrow x=12\,{\rm {ft}}$.
Implicitly taking the derivative of the relation ($x^{2}+y^{2}=r^{2}$) tells us that $2x\cdot {\frac {dx}{dt}}=2r\cdot {\frac {dr}{dt}}$ (notice that the height of the dock is constant). Isolating for ${\frac {dx}{dt}}={\dot {x}}$:
    ${\begin{aligned}{\dot {x}}&={\frac {r}{x}}{\dot {r}}\\&={\frac {13}{5}}\cdot 5\,{\rm {ft/s}}\\&={\textbf {13}}\,{\textbf {ft/s}}\end{aligned}}$
$\sin(\theta )={\frac {x}{r}}={\frac {12}{13}}$
$\cos(\theta )={\frac {5}{13}}$
    ${\begin{aligned}{\dot {\theta }}\cos(\theta )&={\frac {r{\dot {x}}-x{\dot {r}}}{r^{2}}}\\{\dot {\theta }}&={\frac {r{\dot {x}}-x{\dot {r}}}{r^{2}\cos(\theta )}}\\{\dot {\theta }}&={\frac {13\cdot 13-12\cdot 13}{(13)^{2}\cdot {\frac {5}{13}}}}\\&={\frac {1}{5}}={\textbf {0.2}}\,{\textbf {rad/s}}\end{aligned}}$
$\sin(\theta )={\frac {x}{r}}={\frac {12}{13}}$
$\cos(\theta )={\frac {5}{13}}$
    ${\begin{aligned}{\dot {\theta }}\cos(\theta )&={\frac {r{\dot {x}}-x{\dot {r}}}{r^{2}}}\\{\dot {\theta }}&={\frac {r{\dot {x}}-x{\dot {r}}}{r^{2}\cos(\theta )}}\\{\dot {\theta }}&={\frac {13\cdot 13-12\cdot 13}{(13)^{2}\cdot {\frac {5}{13}}}}\\&={\frac {1}{5}}={\textbf {0.2}}\,{\textbf {rad/s}}\end{aligned}}$
$100\,{\rm {m}}$
$2\,{\rm {m}}$
$4\,{\rm {m/s}}$
$\tan(\theta )={\frac {x}{2}}$
    ${\begin{aligned}{\dot {\theta }}\sec ^{2}(\theta )&={\frac {\dot {x}}{2}}\\{\dot {\theta }}&={\frac {\dot {x}}{2\sec ^{2}(\theta )}}\\&={\frac {\dot {x}}{2}}\cos ^{2}(\theta )\end{aligned}}$
Notice that $\theta =\tan ^{-1}\left({\frac {x}{2}}\right)$. Since $x(t)=4t$ after the direct line shot, $x(0.5)=4\cdot 0.5=2$. Therefore, because $\tan ^{-1}\left(1\right)={\frac {\pi }{4}}$,
    ${\begin{aligned}{\dot {\theta }}&={\frac {4}{2}}\cdot \cos ^{2}\left({\frac {\pi }{4}}\right)\\&=2\cdot {\frac {1}{2}}={\textbf {1}}\;{\textbf {rad/s}}\end{aligned}}$
$\tan(\theta )={\frac {x}{2}}$
    ${\begin{aligned}{\dot {\theta }}\sec ^{2}(\theta )&={\frac {\dot {x}}{2}}\\{\dot {\theta }}&={\frac {\dot {x}}{2\sec ^{2}(\theta )}}\\&={\frac {\dot {x}}{2}}\cos ^{2}(\theta )\end{aligned}}$
Notice that $\theta =\tan ^{-1}\left({\frac {x}{2}}\right)$. Since $x(t)=4t$ after the direct line shot, $x(0.5)=4\cdot 0.5=2$. Therefore, because $\tan ^{-1}\left(1\right)={\frac {\pi }{4}}$,
    ${\begin{aligned}{\dot {\theta }}&={\frac {4}{2}}\cdot \cos ^{2}\left({\frac {\pi }{4}}\right)\\&=2\cdot {\frac {1}{2}}={\textbf {1}}\;{\textbf {rad/s}}\end{aligned}}$

## Approximation Problems

By assumption, for these problems, assume $\pi \approx 3.14$ and $e\approx 2.718$ unless stated otherwise. One may use a calculator or design a computer program, but one must indicate the method and reasoning behind every step where necessary.
$\sin(3)$
$f(x,y)={\frac {dy}{dx}}$
$y(x)$
    ${\frac {d}{dx}}\left[\sin(x)\right]=\cos(x)$
We know $\cos(x)={\sqrt {1-\sin ^{2}(x)}}$. Since $y=\sin(x)$,
    $f(x,y)={\sqrt {1-y^{2}}}$.
We will use step size $\Delta x=\pi -3\approx 0.14$ since it will be the easiest to calculate without the use of a calculator. Let $y_{0}=y(\pi )$.
    $y_{1}=0+0.14f\left(\pi ,0\right)$
    $y_{1}=0.14\cdot 1=0.14$
$\sin(3)\approx \mathbf {0.14}$
$f(x,y)={\frac {dy}{dx}}$
$y(x)$
    ${\frac {d}{dx}}\left[\sin(x)\right]=\cos(x)$
We know $\cos(x)={\sqrt {1-\sin ^{2}(x)}}$. Since $y=\sin(x)$,
    $f(x,y)={\sqrt {1-y^{2}}}$.
We will use step size $\Delta x=\pi -3\approx 0.14$ since it will be the easiest to calculate without the use of a calculator. Let $y_{0}=y(\pi )$.
    $y_{1}=0+0.14f\left(\pi ,0\right)$
    $y_{1}=0.14\cdot 1=0.14$
$\sin(3)\approx \mathbf {0.14}$
${\sqrt {2}}$
${\sqrt {2}}$
$f(x)=x^{2}-2$
$f^{\prime }(x)=2x$
$x_{0}=1$
    $x_{1}=1-{\frac {-1}{2}}=1.5$
    $x_{2}=1.5-{\frac {0.25}{3}}\approx 1.4167$
${\sqrt {2}}\approx \mathbf {1.4167}$
${\sqrt {2}}$
$f(x)=x^{2}-2$
$f^{\prime }(x)=2x$
$x_{0}=1$
    $x_{1}=1-{\frac {-1}{2}}=1.5$
    $x_{2}=1.5-{\frac {0.25}{3}}\approx 1.4167$
${\sqrt {2}}\approx \mathbf {1.4167}$
$\ln(3)$
$\ln(e)=1$
$f(x)=\ln(x)$
$f^{\prime }(e)={\frac {1}{e}}$
$(e,1)$
$y={\frac {1}{e}}\left(x-e\right)+1={\frac {x}{e}}$
$3-e$
$y\approx f(x)$
$f(3)\approx {\frac {3}{e}}\approx 1.10$
$\ln(3)\approx \mathbf {1.10} {\text{ OR }}\mathbf {1.09}$
$\ln(e)=1$
$f(x)=\ln(x)$
$f^{\prime }(e)={\frac {1}{e}}$
$(e,1)$
$y={\frac {1}{e}}\left(x-e\right)+1={\frac {x}{e}}$
$3-e$
$y\approx f(x)$
$f(3)\approx {\frac {3}{e}}\approx 1.10$
$\ln(3)\approx \mathbf {1.10} {\text{ OR }}\mathbf {1.09}$

## Advanced Understanding

45. Consider the differentiable function $f(x)$ for all $x>-2$ and continuous function $g(x)$ below, where $g(x)$ is linear for all $x\geq 1$ and differentiable for all $x\in (-2,1)\cup (1,\infty )$, and $f(x)$ and $g(x)$ are continuous for all $x\geq -2$.
$f^{\prime }(1.6)$
    $f^{\prime }(1.6)\approx {\frac {1-3}{2-1.6}}=\mathbf {-5}$
    $f^{\prime }(1.6)\approx {\frac {1-3}{2-1.6}}=\mathbf {-5}$
$\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}$
$\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}\to {\frac {0}{0}}$
$f{\text{ and }}g$
$x=1.6$
$f$
$x\geq -2$
$g$
$x\geq 1$
    ${\begin{aligned}\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&=\lim _{x\to 1.6}{\frac {{\frac {d}{dx}}\left[f(x)-3\right]}{{\frac {d}{dx}}\left[3-g(x)\right]}}\\&=\lim _{x\to 1.6}{\frac {f^{\prime }(x)}{-g^{\prime }(x)}}\end{aligned}}$
Since $g(x)$ is linear for all $x\geq 1$,
    $-g^{\prime }(1.6)=-{\frac {3+4}{1.6-1}}=-{\frac {7}{\frac {3}{5}}}$
Using the answer from (a),
    ${\begin{aligned}\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&={\frac {f^{\prime }(1.6)}{-g^{\prime }(1.6)}}\\&={\frac {-5}{-{\frac {7}{\frac {3}{5}}}}}\\&={\frac {15}{35}}\\\therefore \lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&=\mathbf {\frac {3}{7}} \end{aligned}}$
$\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}\to {\frac {0}{0}}$
$f{\text{ and }}g$
$x=1.6$
$f$
$x\geq -2$
$g$
$x\geq 1$
    ${\begin{aligned}\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&=\lim _{x\to 1.6}{\frac {{\frac {d}{dx}}\left[f(x)-3\right]}{{\frac {d}{dx}}\left[3-g(x)\right]}}\\&=\lim _{x\to 1.6}{\frac {f^{\prime }(x)}{-g^{\prime }(x)}}\end{aligned}}$
Since $g(x)$ is linear for all $x\geq 1$,
    $-g^{\prime }(1.6)=-{\frac {3+4}{1.6-1}}=-{\frac {7}{\frac {3}{5}}}$
Using the answer from (a),
    ${\begin{aligned}\lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&={\frac {f^{\prime }(1.6)}{-g^{\prime }(1.6)}}\\&={\frac {-5}{-{\frac {7}{\frac {3}{5}}}}}\\&={\frac {15}{35}}\\\therefore \lim _{x\to 1.6}{\frac {f(x)-3}{3-g(x)}}&=\mathbf {\frac {3}{7}} \end{aligned}}$
$f^{\prime }(1.6)=f^{\prime }(2)$
$f(x)$
$c\in \mathbb {R}$
$f(c)=0$
$x_{0}=2$
$f(2)=1$
$f^{\prime }(2)=-5$
    ${\begin{aligned}x_{1}=c&=x_{0}-{\frac {f(x_{0})}{f^{\prime }(x_{0})}}\\&=2-{\frac {1}{-5}}=2.2\\\end{aligned}}$
$c=\mathbf {2.2}$
$f(x)$
$c\in \mathbb {R}$
$f(c)=0$
$x_{0}=2$
$f(2)=1$
$f^{\prime }(2)=-5$
    ${\begin{aligned}x_{1}=c&=x_{0}-{\frac {f(x_{0})}{f^{\prime }(x_{0})}}\\&=2-{\frac {1}{-5}}=2.2\\\end{aligned}}$
$c=\mathbf {2.2}$
$f(x)$
$f(x)$
$g(x)$
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