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.

$f\left(x\right)=4-x^2,c=1$

Short answer

The value is$f\text{'}\left(1\right)=-2$.

Step-by-step solution

Step 1. Given Information

We are given a function$f\left(x\right)=4-x^2,c=1$ and using a sequence of approximations we have to estimate $f\text{'}\left(c\right)$.

Step 2. Using a sequence of approximations  

$$\begin{gathered} h=2, \\ \frac{f\left(2\right)-f\left(1\right)}{2-1} \\ =\frac{\left(4-{\left(2\right)}^2\right)-\left(4-{\left(1\right)}^2\right)}{1} \\ =\frac{-3}{1} \\ =-3 \\ h=1.5, \\ \frac{f(1.5)-f\left(1\right)}{1.5-1} \\ =\frac{\left(4-{\left(1.5\right)}^2\right)-\left(4-{\left(1\right)}^2\right)}{0.5} \\ =\frac{1.75-3}{0.5} \\ =-2.5 \\ h=1.1, \\ \frac{f(1.1)-f\left(1\right)}{1.1-1} \\ =\frac{\left(4-{\left(1.1\right)}^2\right)-\left(4-{\left(1\right)}^2\right)}{0.1} \\ =\frac{2.79-3}{0.1} \\ =-2.1 \\ h=0.9, \\ \frac{f(0.9)-f\left(1\right)}{0.9-1} \\ =\frac{\left(4-{\left(0.9\right)}^2\right)-\left(4-{\left(1\right)}^2\right)}{-0.1} \\ =\frac{3.19-3}{-0.1} \\ =-1.9 \end{gathered}$$

We might guess that the slope of the tangent line is $f\text{'}\left(1\right)=-2$.

The graph of the function is

Step 3. Graphs of secant lines

When $c=1,c+h=2$with $h=1$we have $f\left(1\right)=3,f\left(2\right)=0$the secant line can be drawn as follows:

When $c=1,c+h=1.5$with $h=0.5$we have $f\left(1\right)=3,f(1.5)=1.75$the secant line can be drawn as follows:

When $c=1,c+h=1.1$with $h=0.1$we have $f\left(1\right)=3,f(1.1)=2.79$the secant line can be drawn as follows: