Chapter 2.6, Problem 26E

a.

To determine

To find:The equation of the tangent line to the curve $G\left(x\right)=4x^2-x^3$ at the points $\left(2,8\right)\text{ and }\left(3,9\right)$ .

Answer to Problem 26E

The tangent line is $y=-3x+18$ .

Explanation of Solution

Given: $G\left(x\right)=4x^2-x^3$ at the point $\left(2,8\right)\text{ and }\left(3,9\right)$ .

Concept used: $G\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{\left[G\left(x+h\right)-G\left(x\right)\right]}{h}$

ExplanationLet $G\left(x\right)=4x^2-x^3$ be any function.

It has to find the tangent line of the curve and derivative of the function $G\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{\left[G\left(x+h\right)-G\left(x\right)\right]}{h}$

  $\begin{array}{l}G\text{'}\left(a\right)={\underrightarrow{\lim }}_{h\to 0}\frac{\left[4{\left(x+h\right)}^2-{\left(x+h\right)}^3\right]-\left[4x^2-x^3\right]}{h}\\ G\text{'}\left(a\right)={\underrightarrow{\lim }}_{h\to 0}\frac{\left[4\left(x^2+h^2+2xh\right)-\left(x^3+h^3+3x^{2}h+3x{h}^2\right)\right]-\left[4x^2-x^3\right]}{h}\\ G\text{'}\left(a\right)={\underrightarrow{\lim }}_{h\to 0}\frac{\left[4x^2+4h^2+8xh-x^3-h^3-3x^{2}h-3x{h}^2-4x^2+x\right]}{h}\\ ={\underrightarrow{\lim }}_{h\to 0}\frac{\left[4h^2+8xh-h^3-3xh\left(x+h\right)\right]}{h}\\ ={\underrightarrow{\lim }}_{h\to 0}4h+8x-h^2-3x\left(x+h\right)\\ =8x-3x^2\end{array}$

Finding equation of tangent at

  $\left(2,8\right)$

Slope of tangent at

  $\left(2,8\right)$

  $G^{\text{'}}\left(2\right)=8\left(2\right)-3{\left(2\right)}^2=16-12=4$

Therefore the equation of the tangent is

  $\begin{array}{l}y-\left(8\right)=4\left(x-2\right)\\ y-8=\left(4x-8\right)\\ y-4x=0\\ y=4x\end{array}$

Finding equation of tangent at

  $\left(3,9\right)$

Slope of tangent at

  $\left(3,9\right)$

  $G^{\text{'}}\left(3\right)=8\left(3\right)-3{\left(3\right)}^2=24-27=-3$

Therefore the equation of the tangent is

  $\begin{array}{l}y-\left(9\right)=-3\left(x-3\right)\\ y-9=\left(-3x+9\right)\\ y+3x=18\\ y=-3x+18\end{array}$

Hence, the tangent line is $y=-3x+18$ ..

b.

To determine

To illustrate: The graph of the curve and the tangent line on the same screen.

Explanation of Solution

Given: $G\left(x\right)=4x^2-x^3$ at the point $\left(2,8\right)\text{ and }\left(3,9\right)$ .

Let $y=5x/1+x^2$ be any function.

In the graph

The black curve is $y=4x^2-x^3$

The blue line which is tangent at $\left(3,9\right)$ has equation $y=-3x+18$

The red line which is tangent at $\left(2,8\right)$ has equation $y=4x$