Question

For each function f and value $x=c$in Exercises 35–44, use a sequence of approximations to estimate $f\text{'}\left(c\right)$. Illustrate your work with an appropriate sequence of graphs of secant lines.

$f\left(x\right)=4-x^2,c=0$

Short answer

The value is$f\text{'}\left(0\right)=0$.

Step-by-step solution

Step 1. Given Information 

We are given a function $f\left(x\right)=4-x^2,c=0$ and using a sequence of approximations we have to estimate $f\text{'}\left(c\right)$.

Step 2. Using a sequence of approximations   

$$\begin{gathered} h=1, \\ \frac{f\left(1\right)-f\left(0\right)}{1-0} \\ =\frac{\left(4-{\left(1\right)}^2\right)-\left(4-{\left(0\right)}^2\right)}{1} \\ =\frac{3-4}{1} \\ =-1 \\ h=0.5, \\ \frac{f(0.5)-f\left(0\right)}{0.5-0} \\ =\frac{\left(4-{\left(0.5\right)}^2\right)-\left(4-{\left(0\right)}^2\right)}{0.5} \\ =\frac{3.75-4}{0.5} \\ =-0.5 \\ h=0.1, \\ \frac{f(0.1)-f\left(0\right)}{0.1-0} \\ =\frac{\left(4-{\left(0.1\right)}^2\right)-\left(4-{\left(0\right)}^2\right)}{0.1} \\ =\frac{3.99-4}{0.1} \\ =-0.1 \\ h=0.01, \\ \frac{f(0.01)-f\left(0\right)}{0.01-0} \\ =\frac{\left(4-{\left(0.01\right)}^2\right)-\left(4-{\left(0\right)}^2\right)}{0.01} \\ =-0.01 \end{gathered}$$

We might guess that the slope of the tangent line is $f\text{'}\left(0\right)=0$.

The graph of the function is

Step 3. Graphs of secant lines 

When $c=0,c+h=1$with $h=1$we have $f\left(0\right)=4,f\left(1\right)=3$the secant line can be drawn as follows:

When $c=0,c+h=0.5$with $h=0.5$we have $f\left(0\right)=4,f(0.5)=3.75$the secant line can be drawn as follows:

When $c=0,c+h=0.1$with $h=0.1$we have $f\left(0\right)=4,f(0.1)=3.99$the secant line can be drawn as follows: