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Answers

Math V46 Section 2.5

Introduction

Please try this problem on your own before looking at my solution. There may be other ways to solve this; I make no claim that this method is the quickest, prettiest, or most efficient. Also, I did not look at the student solutions manual, so this approach may or may not match what they did.

Problem 60

The given function is $$f(x) = {x^4} - 32{x^2} + 10$$

(A)

$\displaystyle{f'(x) = \frac{d}{{dx}}\left( {{x^4} - 32{x^2} + 10} \right)}$

$\displaystyle{f'(x) = \frac{d}{{dx}}\left( {{x^4}} \right) - \frac{d}{{dx}}\left( {32{x^2}} \right) + \frac{d}{{dx}}\left( {10} \right)\hspace{3em}{\text{(justification: sum property)}}}$

$\displaystyle{f'(x) = \frac{d}{{dx}}\left( {{x^4}} \right) - 32\frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( {10} \right)\hspace{3em}{\text{(justification: constant multiple property)}}}$

$\displaystyle{f'(x) = 4{x^3} - 32(2x) + 0 \hspace{3em}{\text{(justification: power rule, constant function rule)}}}$ $$\boxed{\displaystyle{f'(x) = 4{x^3} - 64x}}$$

(B)

Slope of the graph at $x=2$: $$\text{slope} = f'(2) = 4{[(2)^3]} - 64(2) = 4(8) - 128 = 32 - 128 = \boxed{\displaystyle{-96}}$$

Slope of the graph at $x=4$: $$\text{slope} = f'(4) = 4{[(4)^3]} - 64(4) = 4(64) - 256 = 256 - 256 = \boxed{\displaystyle{0}}$$

(C)

We have already obtained the slopes of the tangent lines at $x=2$ and $x=4$ in part (B) above. To obtain the equations of the tangent lines, we will repeatedly employ the point-slope formula $$y - {y_1} = m\left( {x - {x_1}} \right)$$

At $x=2$, we will use $m=-96$ (from the first slope result in part (B)) and $x_1 = 2$ (since this is the given value of $x$). To obtain $y_1$, we must plug $x_1 = 2$ into the original function $f(x)$ to obtain the corresponding $y$ value. The necessary computation is $${y_1} = f(2) = {\left( 2 \right)^4} - 32{\left( 2 \right)^2} + 10 = 16 - 32\left( 4 \right) + 10 = 16 - 128 + 10 = \underline{\underline { - 102}}$$ Note that the above calculation determines $f(2)$, not $f'(2)$. Inserting the values for $m$, $x_1$, and $y_1$ into the point-slope formula yields $$y - (- 102) = (-96)\left( {x - (2)} \right)$$ Simplifying and isolating $y$ gives the final slope-intercept form of the answer: $$y + 102 = -96x + 192$$ $$\boxed{\displaystyle{y = -96x + 90}}$$

At $x=4$, we will use $m=0$ (from the second slope result in part (B)) and $x_1 = 4$ (since this is the given value of $x$). To obtain $y_1$, we must plug $x_1 = 4$ into the original function $f(x)$ to obtain the corresponding $y$ value. The necessary computation is $${y_1} = f(4) = {\left( 4 \right)^4} - 32{\left( 4 \right)^2} + 10 = 256 - 32\left( 16 \right) + 10 = 256 - 512 + 10 = \underline{\underline { - 246}}$$ Note that the above calculation determines $f(4)$, not $f'(4)$. Inserting the values for $m$, $x_1$, and $y_1$ into the point-slope formula yields $$y - (- 246) = (0)\left( {x - (4)} \right)$$ Simplifying and isolating $y$ gives the final slope-intercept form of the answer: $$y + 246 = 0$$ $$\boxed{\displaystyle{y = -246}}$$ This latter result represents a horizontal line, since the slope of the tangent at $x=4$ is zero.

(D)

Here we are asked to determine the values of $x$ at which the tangent line is horizontal (that is, has a slope of zero). We have already accidentally found one such point at $x=4$ (see the second result in part (B) above). To find the remaining values of $x$, we note that if the tangent line is horizontal, then the slope (and therefore the derivative) has a value of zero. We can force this to happen by simply setting the general expression for $f'(x)$ equal to zero, then using algebraic methods to isolate $x$. The system of equations to solve is therefore

$f'(x) = 0$ and $f'(x) = 4{x^3} - 64x$ (the former is the condition for a horizontal tangent, and the latter is obtained from part (A)).

Combining these equations gives $$4{x^3} - 64x = 0$$ which can be solved by repeated factoring: $$4x \left( x^2 - 16 \right) = 0$$ $$4x \left( x + 4 \right) \left( x - 4 \right) = 0$$ So either $4x = 0$, $\left( x + 4 \right) = 0$, or $\left( x - 4 \right) = 0$. The solutions are $$\boxed{\displaystyle{x = 0, x = - 4, x = 4}}$$