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.

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

Short answer

We have approximated the slope by using the concept of the secant line.

Step-by-step solution

Step 1. Given information.

We have to use a sequence of approximations to estimate$f^{\text{'}}\left(c\right)$

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

Step 2. Use sequence of approximation

Let,

$h=2,1.5,1.25,1.1$

Consider the expressions,

$\begin{array}{r}\frac{f\left(2\right)-f\left(1\right)}{2-1}=\frac{\left[0\right]-\left[3\right]}{1}\\ =-3\\ \frac{f\left(1.5\right)-f\left(1\right)}{1.5-1}=\frac{\left[\left|(1.5{)}^2-4\right|\right]-\left[3\right]}{0.5}\\ =-2.5\end{array}$

and,

$\begin{array}{r}\frac{f\left(1.25\right)-f\left(1\right)}{1.25-1}=\frac{\left[\left|(1.25{)}^2-4\right|\right]-\left[3\right]}{0.25}\\ =-2.25\\ \frac{f\left(1.1\right)-f\left(1\right)}{1.1-1}=\frac{\left[\left|(1.1{)}^2-4\right|\right]-\left[3\right]}{0.1}\\ =-2.1\end{array}$

The slope of the tangent will be :

$f^{\mathrm{\prime }}\left(1\right)=-2$

The graph is ;

Step 3. First secant graph

Take c=1 , c+h=2, then the corresponding values are:

$f\left(1\right)=3,f\left(2\right)=0$

The secant line can be drawn as:

Step 4. Second secant graph

Take c=3 and c+h = 3.5 then the corresponding values are :

$f\left(3\right)=2,f\left(3.5\right)=2.5$

The secant graph is :

Step 5. Third secant graph

Take c=1 and c+h=1.5, then the corresponding values are :

$f\left(1\right)=3,f\left(1.5\right)=1.75$

The secant graph is :