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Expert-verified Found in: Page 777 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # 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. role="math" localid="1648705352169" $f\left(x\right)={e}^{x},c=0$

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

See the step by step solution

## Step 1. Given information.

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

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

## 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 :  ### Want to see more solutions like these? 