Chapter 3.2, Problem 22E

To determine

The numerical derivative of the given function at the indicate point. Also, identify the function is differentiable at the indicated point or not.

Answer to Problem 22E

The given function $f\left(x\right)$ is differentiable at the indicated point and the value of the function is $8$ .

Explanation of Solution

Given information:

The given function: $f\left(x\right)=x^3-4x,x=2$

The value of $h=0.001$

Formula Used:

Differentiable function formula:

  $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x-h\right)}{2h}$

Calculation:

The given function is $f\left(x\right)=x^3-4x$ and $x=2$ .

To find the numerical derivative of the given function at the indicate point, use the differentiable function formula $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x-h\right)}{2h}$ .

Now, find the value of $f\left(x+h\right)$ and $f\left(x-h\right)$ using the given function.

  $f\left(x+h\right)={\left(x+h\right)}^3-4\left(x+h\right)$ and $f\left(x-h\right)={\left(x-h\right)}^3-4\left(x-h\right)$ .

Substitute $0.001$ for $h$ and $2$ for $x$ in the equation $f\left(x+h\right)={\left(x+h\right)}^3-4\left(x+h\right)$ .

  $\begin{array}{c}f\left(x+h\right)={\left(2+0.001\right)}^3-4\left(2+0.001\right)\\ ={\left(2.001\right)}^3-4\left(2.001\right)\\ =8.012006001-8.004\\ =0.008006001\end{array}$

Also, substitute $0.001$ for $h$ and $2$ for $x$ in the equation $f\left(x-h\right)={\left(x-h\right)}^3-4\left(x-h\right)$ .

  $\begin{array}{c}f\left(x-h\right)={\left(2-0.001\right)}^3-4\left(2-0.001\right)\\ ={\left(1.999\right)}^3-4\left(1.999\right)\\ =7.988005999-7.996\\ =-0.007994001\end{array}$

Now, substitute $0.008006001$ for $f\left(x+h\right)$ , $-0.007994001$ for $f\left(x-h\right)$ and $0.001$ for $h$ in the formula $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x-h\right)}{2h}$ .

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x-h\right)}{2h}=\underset{h\to 0}{\lim }\frac{\left(0.008006001\right)-\left(-0.007994001\right)}{2\left(0.001\right)}\\ =\frac{0.008006001+0.007994001}{0.002}\\ =\frac{0.016000002}{0.002}\\ =8.000001\\ \approx 8\end{array}$

Thus, the given function $f\left(x\right)$ is a polynomial function. Also, it is continuous and differentiable on its domain.

Hence, the given function $f\left(x\right)$ is differentiable at the indicated point and the required value of the function is $8$ .