Chapter 2.6, Problem 9E

(a)

To determine

To find: The slope of the tangent line to the curve.

Answer to Problem 9E

The slope of the tangent line to the curve $y=3+4x^2-2x^3$ at the point where

$x=a$ is $m=8a-6a^2$.

Explanation of Solution

Given:

The equation of the curve is $y=3+4x^2-2x^3$.

The curve passing through the points (1, 5) and (2, 3).

Formula used:

The slope of the tangent curve $y=f\left(x\right)$ at the point $P\left(a,f\left(a\right)\right)$ is,

$m=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ (1)

Calculation:

Obtain the slope of the tangent to the curve at the point $x=a$.

Since $f\left(x\right)=3+4x^2-2x^3$, substitute $f\left(a\right)=3+4a^2-2a^3$ and $f\left(a+h\right)=3+4{\left(a+h\right)}^2-2{\left(a+h\right)}^3$ in equation (1),

$\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(3+4{\left(a+h\right)}^2-2{\left(a+h\right)}^3\right)-\left(3+4a^2-2a^3\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(3+4\left(a^2+h^2+2ah\right)-2\left(a^3+3a^{2}h+3a{h}^2+h^3\right)\right)-\left(3+4a^2-2a^3\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(3+4a^2+4h^2+4\cdot 2ah-2a^3-2\cdot 3a^{2}h-2\cdot 3a{h}^2-2h^3\right)-\left(3+4a^2-2a^3\right)}{h}\end{array}$

Perform the mathematical operations as shown below.

$\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{3+4a^2+4h^2+8ah-2a^3-6a^{2}h-6a{h}^2-2h^3-3-4a^2+2a^3}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(3-3\right)+\left(4a^2-4a^2\right)+\left(2a^3-2a^3\right)+4h^2+8ah-6a^{2}h-6a{h}^2-2h^3}{h}\\ =\underset{h\to 0}{\lim }\frac{4h^2+8ah-6a^{2}h-6a{h}^2-2h^3}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(4h+8a-6a^2-6ah-2h^2\right)}{h}\end{array}$

Since the limit h approaches 0 but not equal to 1, cancel the common term $h\left(\ne 0\right)$ from both the numerator and the denominator,

$\begin{array}{c}m=\underset{h\to 0}{\lim }\left(4h+8a-6a^2-6ah-2h^2\right)\\ =\left(4\left(0\right)+8a-6a^2-6a\left(0\right)-2{\left(0\right)}^2\right)\\ =0+8a-6a^2-0-0\\ =8a-6a^2\end{array}$

Thus, the slope of the tangent line to the curve at the point $x=a$ is $m=8a-6a^2$.

(b)

To determine

To find: The equation of the tangent lines to the curve $y=3+4x^2-2x^3$ at the points $\left(1,5\right)$ and (2, 3).

Answer to Problem 9E

The equation of the tangent lines to the curve $y=3+4x^2-2x^3$ at the points (1, 5) and

(2, 3) are $y=2x+3$ and $y=-8x+19$ respectively.

Explanation of Solution

Formula used:

The equation of the tangent line to the curve $y=f\left(x\right)$ at the point $\left(a,f\left(a\right)\right)$ is,

$y-f\left(a\right)=f^{\prime }\left(a\right)\left(x-a\right)$ (2)

Since the tangent line to the curve $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is the line through the point $\left(a,f\left(a\right)\right)$ whose slope is equal to the derivative of a, the value of $f^{\prime }\left(a\right)=8a-6a^2$.

At the point (1, 5), take $a=1$ and $f\left(a\right)=5$

$\begin{array}{c}{f}^{\prime }\left(1\right)=8\left(1\right)-6{\left(1\right)}^2\\ =8-6\\ =2\end{array}$

Substitute $a=1$, $f\left(a\right)=5$ and $f^{\prime }\left(a\right)=2$ in equation (2),

$\begin{array}{c}\left(y-f\left(a\right)\right)=f^{\prime }\left(a\right)\left(x-a\right)\\ y-5=2\left(x-1\right)\\ y-5=2x-2\end{array}$

Isolate y as shown below:

$\begin{array}{c}y=2x-2+5\\ =2x+3\end{array}$

Thus, the equation of the tangent line is $y=2x+3$.

Obtain the equation of the tangent line at the point (2, 3).

Since the tangent line to the curve $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is the line through the point $\left(a,f\left(a\right)\right)$ whose slope is equal to the derivative of a, the value of $f^{\prime }\left(a\right)=8a-6a^2$.

At the point (2, 3), take $a=2$ and $f\left(a\right)=3$

$\begin{array}{c}{f}^{\prime }\left(2\right)=8\left(2\right)-6{\left(2\right)}^2\\ =16-24\\ =-8\end{array}$

Substitute $a=2$, $f\left(a\right)=3$ and $f^{\prime }\left(a\right)=-8$ in equation (2).

$\begin{array}{c}\left(y-f\left(a\right)\right)=f^{\prime }\left(a\right)\left(x-a\right)\\ y-3=-8\left(x-2\right)\\ y-3=-8x+16\end{array}$

Isolate y as shown below:

$\begin{array}{c}y=-8x+16+3\\ =-8x+19\end{array}$

Thus, the equation of the tangent line is $y=-8x+19$.

(c)

To determine

To sketch: The graph of the curve and the tangent lines.

Explanation of Solution

Given:

The equation of the curve is $y=3+4x^2-2x^3$.

The equation of the tangent lines are $y=2x+3$ and $y=-8x+19$.

Graph:

Use the online graphing calculator to draw the graph of the functions as shown below in Figure 1.

Notice that, the two lines $y=2x+3$ and $y=-8x+19$ touches on the curve $y=3+4x^2-2x^3$ at a unique point. That is, the two lines are tangent to the curve $y=3+4x^2-2x^3$.