We can also conclude from the information in the table that f is not continuous at 2 or 3 or 4, because f(x) ≠ f(2) , f(x) ≠ f(3) , and f(x) ≠ f(4). The behaviors at x = 2 and x = 4 exhibit a hole in the graph, sometimes called a removable discontinuity, since the graph could be made continuous by changing the value of a single point. The behavior at x = 3 is called a jump discontinuity, since the graph jumps between two values. So which functions are continuous? It turns out pretty much every function you’ve studied is continuous where it is defined: polynomial, radical, rational, exponential, and logarithmic functions are all continuous where they are defined. Moreover, any combination of continuous functions is also continuous. This is helpful, because the definition of continuity says that for a continuous function, . That means for a continuous function, we can find the limit by direct substitution (evaluating the function) if the function is continuous at a. Example 4 Evaluate using continuity, if possible: a) b) b) a) The given function is polynomial, and is defined for all values of x, so we can find the limit by direct substitution: b) The given function is rational. It is not defined at x = -3, but we are taking the limit as x approaches 2, and the function is defined at that point, so we can use direct substitution: c) This function is not defined at x = 2, and so is not continuous at x = 2. We cannot use direct substitution. 2.2 Exercises 1. Use the graph to determine the following limits. (a) (b) (c) (d) 2. Use the graph to determine the following limits. (a) (b)
are given. Use the Second Derivative Test to determine whether each point (x, f(x)) is a local maximum, a local minimum or neither 22. f(x) = 2x3 – 15x2 + 6 , x = 0 , 5 . 23. g(x) = x3 – 3x2 – 9x + 7 , x = –1 , 3 . 24. h(x) = x4 – 8x2 – 2 , x = –2, 0, 2 .
Solving for y', we have , and, at the point (3,4), y' = – 3/4. In the previous example, it would have been easy to explicitly solve for y, and then we could differentiate y to get y '. Because we could explicitly solve for y , we had a choice of methods for calculating y '. Sometimes, however, we can not explicitly solve for y , and the only way of determining y ' is implicit differentiation. Related Rates If several variables or quantities are related to each and some of the variables are changing at a known rate, then we can use derivatives to determine how rapidly the other variables must be changing. Example 3 Suppose the border of a town is roughly circular, and the radius of that circle has been increasing at a rate of 0.1 miles each year. Find how fast the area of the town has been increasing when the radius is 5 miles. We could get an approximate answer by calculating the area of the circle when the radius is 5 miles ( A = πr2 = π(5 miles)2 ≈ 78.6 miles2 ) and 1 year later when the radius is 0.1 feet larger than before ( A = πr2 = π(5.1 miles)2 ≈ 81.7 miles2 ) and then finding ∆Area/∆time = (81.7 mi2 – 78.6 mi2)/(1 year) = 3.1 mi2/yr. This approximate answer represents the average change in area during the 1 year period when the radius increased from 5 miles to 5.1 miles, and would correspond to the secant slope on the area graph. To find the exact answer, though, we need derivatives. In this case both radius and area are functions of time: r(t) = radius at time t A(t) = area at time t We know how fast the radius is changing, is a statement about the derivative: . We also know that r = 5 at our moment of interest. We are looking for how fast the area is increasing, which is . Now we need an equation relating our variables, which is the area equation: . Taking the derivative of both sides of that equation with respect to t, we can use implicit differentiation: Plugging in the values we know for r and dr/dt,