Q. 40 - Calculus

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.

$f\left(x\right)=e^x,c=0$

Step-by-Step Solution

Answer

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

Step 1. Given information.

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

$f\left(x\right)=e^x,c=0$

Step 2. Use sequence of approximation

Let,

$h=1,0.5,0.1,0.01$

Consider the expressions,

$\begin{array}{r}\frac{f\left(1\right)-f\left(0\right)}{1-0}=\frac{\left[e\right]-1}{1}\\ =1.718\\ \frac{f\left(0.5\right)-f\left(0\right)}{0.5-0}=\frac{\left[e^{0.5}\right]-1}{0.5}\\ =1.297\\ \frac{f\left(0.1\right)-f\left(0\right)}{1.1-0}=\frac{\left[e^{0.1}\right]-\left[1\right]}{0.1}\\ =1.051\\ \frac{f\left(0.01\right)-f\left(0\right)}{0.01-0}=\frac{\left[e^{0.01}\right]-\left[1\right]}{0.01}\\ =1.005\end{array}$

The slope of tangent will be :

$f^{\mathrm{\prime }}\left(0\right)=1$

The graph is ;

Step 3. First secant graph

Take c=0 , c+h=1, then the corresponding values are:

$f\left(0\right)=1,f\left(1\right)=2.718$

The secant line can be drawn as:

Step 4. Second secant graph

Take c=0 and c+h = 0.1 then the corresponding values are :

$f\left(0\right)=1,f\left(0.1\right)=1.1051$

The secant graph is :

Step 5. Third secant graph

Take c=0 and c+h=0.01, then the corresponding values are : 

$f\left(0\right)=1,f\left(0.01\right)=1.0101$

The secant graph is :