user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D106(%rip),%rdi
  lea    D8(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x38,%rsp
  call   <puts@plt>
  lea    D48(%rip),%rdi
  call   <puts@plt>
  pxor   %xmm2,%xmm2
  movapd %xmm2,%xmm0
  call   <log@plt>
  pxor   %xmm4,%xmm4
  movsd  %xmm0,0x8(%rsp)
  movapd %xmm4,%xmm0
  call   <f@plt>
  pxor   %xmm2,%xmm2
  xor    %esi,%esi
  mov    $0x4,%eax
  movapd %xmm0,%xmm1
  movapd %xmm2,%xmm0
  lea    D73(%rip),%rdi
  movsd  0x8(%rsp),%xmm3
  divsd  D113(%rip),%xmm3
  call   <printf@plt>
  movsd  D58(%rip),%xmm2
  movsd  D80(%rip),%xmm4
  movsd  D118(%rip),%xmm5
  nopl   0x0(%rax,%rax,1)
L24:
  movapd %xmm4,%xmm0
  movsd  %xmm2,0x28(%rsp)
  movsd  %xmm4,0x10(%rsp)
  movsd  %xmm5,0x8(%rsp)
  call   <f@plt>
  movsd  0x8(%rsp),%xmm5
  movapd %xmm0,%xmm3
  mulsd  %xmm5,%xmm3
  movapd %xmm5,%xmm0
  movsd  %xmm5,0x20(%rsp)
  movsd  %xmm3,0x8(%rsp)
  call   <f@plt>
  movsd  0x10(%rsp),%xmm4
  movsd  0x8(%rsp),%xmm3
  mulsd  %xmm4,%xmm0
  subsd  %xmm0,%xmm3
  movapd %xmm4,%xmm0
  movsd  %xmm3,0x18(%rsp)
  call   <f@plt>
  movsd  0x20(%rsp),%xmm5
  movsd  %xmm0,0x8(%rsp)
  movapd %xmm5,%xmm0
  call   <f@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  0x18(%rsp),%xmm3
  movsd  0x28(%rsp),%xmm2
  movsd  0x10(%rsp),%xmm4
  subsd  %xmm0,%xmm1
  movsd  %xmm2,0x10(%rsp)
  movsd  %xmm4,0x20(%rsp)
  divsd  %xmm1,%xmm3
  movapd %xmm3,%xmm2
  movsd  %xmm3,0x28(%rsp)
  subsd  %xmm4,%xmm2
  andpd  D54(%rip),%xmm2
  movapd %xmm2,%xmm0
  movsd  %xmm2,0x18(%rsp)
  call   <log@plt>
  movsd  0x10(%rsp),%xmm1
  movsd  %xmm0,0x8(%rsp)
  movapd %xmm1,%xmm0
  call   <log@plt>
  movsd  0x8(%rsp),%xmm3
  movsd  0x28(%rsp),%xmm4
  divsd  %xmm0,%xmm3
  movapd %xmm4,%xmm0
  movsd  %xmm4,0x8(%rsp)
  movsd  %xmm3,0x10(%rsp)
  call   <f@plt>
  movsd  0x8(%rsp),%xmm4
  mov    %ebx,%esi
  mov    %rbp,%rdi
  movsd  0x18(%rsp),%xmm2
  movsd  0x10(%rsp),%xmm3
  movapd %xmm0,%xmm1
  add    $0x1,%ebx
  movapd %xmm4,%xmm0
  mov    $0x4,%eax
  movsd  %xmm2,0x10(%rsp)
  call   <printf@plt>
  movsd  0x8(%rsp),%xmm4
  movapd %xmm4,%xmm0
  call   <f@plt>
  pxor   %xmm7,%xmm7
  movsd  0x8(%rsp),%xmm4
  movsd  0x10(%rsp),%xmm2
  ucomisd %xmm7,%xmm0
  movsd  0x20(%rsp),%xmm5
  jp     L24
  jne    L24
  movapd %xmm4,%xmm0
  subsd  %xmm4,%xmm0
  andpd  D54(%rip),%xmm0
  comisd D91(%rip),%xmm0
  ja     L24
  movapd %xmm4,%xmm0
  lea    D22(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D106
data_type: string
name: parse_data
arguments: label: D8
data_type: string
name: parse_data
arguments: label: D48
data_type: string
name: parse_data
arguments: label: D73
data_type: string
name: parse_data
arguments: label: D113
data_type: byte[8]
name: parse_data
arguments: label: D58
data_type: f64
name: parse_data
arguments: label: D80
data_type: f64
name: parse_data
arguments: label: D118
data_type: f64
name: parse_data
arguments: label: D54
data_type: qword[2]
name: parse_data
arguments: label: D91
data_type: f64[2]
name: parse_data
arguments: label: D22
data_type: string

tool: D106: Solution of e^x = 2x +1 using secants method

tool: D8: %d	%f	%f	%f	%f


tool: D48: Initial guess: 1

tool: D73: i	Root		f(Root)		Error		order of convergence


tool: {"D113": [0xef,0x39,0xfa,0xfe,0x42,0x2e,0xe6,0x3f]}

tool: D58: 2

tool: D80: 3

tool: D118: 1

tool: {"D54": [0x7fffffffffffffff,0x0000000000000000]}

tool: {"D91": [1e-6,null]}

tool: D22: 

Final Answer:%f


assistant: ```c
int main() {
  printf("Solution of e^x = 2x +1 using secants method\n");
  printf("Initial guess: 1\n");
  double veryOldRoot = 1, oldRoot = 3, newRoot;
  double oldError = 2;
  double newError;
  int i = 0;
  printf("i\tRoot\t\tf(Root)\t\tError\t\torder of convergence\n", i++, newRoot, f(newRoot), newError, log(newError) / log(oldError));
  do {
    newRoot = ((veryOldRoot * f(oldRoot) - oldRoot * f(veryOldRoot)) / (f(oldRoot) - f(veryOldRoot)));
    newError = fabs(newRoot - oldRoot);
    printf("%d\t%f\t%f\t%f\t%f\n", i++, newRoot, f(newRoot), newError, log(newError) / log(oldError));
    veryOldRoot = oldRoot;
    oldRoot = newRoot;
    oldError = newError;
  } while ((f(newRoot) != 0) || (fabs(newRoot - oldRoot) > 0.000001));
  printf("\n\nFinal Answer:%f\n", newRoot);
  return 0;
}
```