Q. 20 - Calculus

Question

The derivative of a smooth function f at a point $x=c$ can also be approximated with a symmetric difference quotient:

$f\text{'}\left(c\right)\approx \frac{f(c+h)-f(c-h)}{2h}$

(a) Use a graph to illustrate what the symmetric difference measures. Why would it be reasonable to use the two-sided symmetric difference to approximate f'(c)? (Hint: Your answer should involve a certain kind of secant line and a discussion of what happens as h gets close to $0$.)

(b) Use a sequence of symmetric difference approximations to estimate the derivative of $f\left(x\right)=x^2 at x=3$ Illustrate your answer with a sequence of graphs.

Step-by-Step Solution

Answer

(a) Symmetric function is used to find the derivative of function at the point. 

(b) From the graph, as the slope of the secant line approaches the tangent lien, the secant line overlaps the tangent line.

As the value of h approaches $0$, the derivative of the function at $x=3$ approaches six.

Part (a) Step 1. Given Information.

The function can be approximated with $f\text{'}\left(c\right)\approx \frac{f(c+h)-f(c-h)}{2h}$

Part (a) Step 2. Find the value of c for h = 1 . 0 . 5 , 0 . 1

Consider the function $f\left(x\right)=x^3$

Assume $h=1,0.5,0.1$ and $c=3$

Now, substitute: 

$$\begin{gathered} c=1, \\ h=1 \end{gathered}$$

$$\begin{gathered} f\text{'}\left(1\right)=\frac{f(1+1)-f\left(1\right)}{2\left(1\right)} \\ =\frac{2^3-0^3}{2} \\ =4 \end{gathered}$$

Now substitute:

$$\begin{gathered} c=1, \\ h=0.5 \end{gathered}$$

$$\begin{gathered} f\text{'}\left(1\right)=\frac{f(1+0.5)-f(0.5)}{2(0.5)} \\ =\frac{{(1.5)}^3-{(0.5)}^3}{1} \\ =3.25 \end{gathered}$$

Part (a) Step 3. Find the derivative of c for h = 0 . 1

Substitute $c=1,h=0.1$

$$\begin{gathered} f\text{'}\left(1\right)=\frac{f(1+0.1)-f(0.1)}{2(0.1)} \\ =\frac{{(1.1)}^3-{(0.9)}^3}{0.2} \\ =3.01 \end{gathered}$$

From the above consideration, we can conclude that the value of $f\text{'}\left(1\right)$ approaches $c=3$ when the value of $h$approaches zero.

Part (a) Step 4. Find slope at ( 1 , 1 )

By using slope formula through the point $(1,1)$

$$\begin{gathered} y-1=3(x-1) \\ y-1=3x-3 \\ y=3x-2 \end{gathered}$$

The tangent line at the point $(1,1)$ is $y=3x-2$.

Consider the first secant line with slope $4$ that passes through $(1,1)$

$$\begin{gathered} y-1=4(x-1) \\ y-1=4x-4 \\ y=4x-3 \end{gathered}$$

Part (a) Step 5. Find secant lines for the obtained slopes.

Consider the second secant line with slope $3.25$ that passes through $(1,1)$

$$\begin{gathered} y-1=3.25(x-1) \\ y-1=3.25x-3.25 \\ y=3.25x-2.25 \end{gathered}$$

Consider the third secant line with slope $3.01$ that passes through $(1,1)$.

$$\begin{gathered} y-1=3.01(x-1) \\ y-1=3.01x-3.01 \\ y=3.01x-2.01 \end{gathered}$$

Part (a) Step 6. Graph the observed function with tangent lines.

Graph the function with its secant lines and tangent lines on the same coordinate axis.

From the figure, as the slope of the secant line approaches the slope of the tangent line, so the slope of the secant line overlaps the tangent line.

So when the value of h approaches zero, the derivative of function at $x=1$ approaches 3

Therefore symmetric function is used to find the derivative of function at the point.

Part (b) Step. 1 Find f ' ( c ) for the function.

$f\left(x\right)=x^2$

$$\begin{gathered} f(c+h)={(c+h)}^2 \\ =c^2+2ch+h^2 \\ f(c-h)=c^2-2ch+h^2 \end{gathered}$$

$$\begin{gathered} f\text{'}\left(c\right)\approx \frac{(c^2+2ch+h^2)-(c^2+2ch+h^2)}{2h} \\ =\frac{4ch}{2h} \\ =2c \end{gathered}$$

Part (b) Step 2. Find the derivative.

To find the value of $f\text{'}\left(3\right)$

$$\begin{gathered} f\text{'}\left(3\right)=2\left(3\right) \\ =6 \end{gathered}$$

Part (b) Step 3. Find the tangent line.

To find the tangent at the point $(3,9)$,

$$\begin{gathered} y-9=6(x-3) \\ y-9=6x-18 \\ y=6x-9 \end{gathered}$$

Part (b) Step 4. Find the secant lines.

Consider the first secant line with slope $7$ that passes through $(3,9)$,

$$\begin{gathered} y-9=7(x-3) \\ y-9=7x-21 \\ y=7x-12 \end{gathered}$$

Consider the second secant line with slope $6.5$ that passes through $(3,9)$,

$$\begin{gathered} y-9=6.5(x-3) \\ y-9=6.5x-19.5 \\ y=6.5x-10.5 \end{gathered}$$

Consider the third secant line with slope $6.1$ that passes through $(3,9)$,

$$\begin{gathered} y-9=6.1(x-3) \\ y-9=6.1x-18.3 \\ y=6.1x-9.3 \end{gathered}$$

Part (b) Step 5. Graph the function.

Graph the function with secant line and tangent line.

Part (b) Step 6. Observe the result from the graph.

From the graph, as the slope of the secant line approaches the tangent lien, the secant line overlaps the tangent line.

As the value of h approaches $0$, the derivative of the function at $x=3$ approaches six.