Q. 38 - 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.

$f\left(x\right)=x+x^3, c=1$

Step-by-Step Solution

Answer

The value is $f\text{'}\left(1\right)=4$. 

Step 1. Given Information

We are given a function $f\left(x\right)=x+x^3, c=1$ and using a sequence of approximations we have to estimate $f\text{'}\left(c\right)$.  

Step 2. Using a sequence of approximations

$$\begin{gathered} h=2, \\ \frac{f\left(2\right)-f\left(1\right)}{2-1} \\ =\frac{\left(2+{\left(2\right)}^3\right)-\left(1+{\left(1\right)}^3\right)}{1} \\ =\frac{10-2}{1} \\ =8 \\ h=1.5, \\ \frac{f(1.5)-f\left(1\right)}{1.5-1} \\ =\frac{\left(1.5+{\left(1.5\right)}^3\right)-\left(1+{\left(1\right)}^3\right)}{0.5} \\ =\frac{4.875-2}{0.5} \\ =5.75 \\ h=1.1, \\ \frac{f(1.1)-f\left(1\right)}{1.1-1} \\ =\frac{\left(1.1+{\left(1.1\right)}^3\right)-\left(1+{\left(1\right)}^3\right)}{0.1} \\ =\frac{2.431-2}{0.1} \\ =4.31 \\ h=0.9, \\ \frac{f(0.9)-f\left(1\right)}{0.9-1} \\ =\frac{\left(0.9+{\left(0.9\right)}^3\right)-\left(1+{\left(1\right)}^3\right)}{-0.1} \\ =\frac{1.629-2}{-0.1} \\ =3.71 \end{gathered}$$

We might guess that the slope of the tangent line is $f\text{'}\left(1\right)=4$.

The graph of the function is 

Step 3. Graphs of secant lines

When $c=1, c+h=2$ with $h=1$ we have $f\left(1\right)=2, f\left(2\right)=10$ the secant line can be drawn as follows:

When $c=1, c+h=1.5$ with $h=0.5$ we have $f\left(1\right)=2, f(1.5)=4.875$ the secant line can be drawn as follows:

When $c=1, c+h=1.1$ with $h=0.1$ we have $f\left(1\right)=2, f(1.1)=2.431$ the secant line can be drawn as follows: