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.

$f\left(x\right)=x+x^3,c=0$

Short answer

The value is$f\text{'}\left(0\right)=1$.

Step-by-step solution

Step 1. Given Information  

We are given a function $f\left(x\right)=x+x^3,c=0$ and using a sequence of approximations we have to estimate $f\text{'}\left(c\right)$.

Step 2. Using a sequence of approximations    

$$\begin{gathered} h=1, \\ \frac{f\left(1\right)-f\left(0\right)}{1-0} \\ =\frac{\left(1+{\left(1\right)}^3\right)-\left(0+{\left(0\right)}^3\right)}{1} \\ =\frac{2-0}{1} \\ =2 \\ h=0.5, \\ \frac{f(0.5)-f\left(0\right)}{0.5-0} \\ =\frac{\left(0.5+{\left(0.5\right)}^3\right)-\left(0+{\left(0\right)}^3\right)}{0.5} \\ =\frac{0.625-0}{0.5} \\ =1.25 \\ h=0.1, \\ \frac{f(0.1)-f\left(0\right)}{0.1-0} \\ =\frac{\left(0.1+{\left(0.1\right)}^3\right)-\left(0+{\left(0\right)}^3\right)}{0.1} \\ =\frac{0.101-0}{0.1} \\ =1.01 \\ h=0.01, \\ \frac{f(0.01)-f\left(0\right)}{0.01-0} \\ =\frac{\left(0.01+{\left(0.01\right)}^3\right)-\left(0+{\left(0\right)}^3\right)}{0.01} \\ =\frac{0.010001}{0.01} \\ =1.0001 \end{gathered}$$

We might guess that the slope of the tangent line is $f\text{'}\left(0\right)=1$.

The graph of the function is

Step 3. Graphs of secant lines  

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

When $c=0,c+h=0.5$with $h=0.5$we have $f\left(0\right)=0,f(0.5)=0.625$the secant line can be drawn as follows:

When $c=0,c+h=0.1$with $h=0.1$we have $f\left(0\right)=0,f(0.1)=0.101$the secant line can be drawn as follows: