Single Variable Calculus

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Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 8

Expert Answers

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Use the guidelines of curve sketching to sketch the curve. $y =(4-x^2)^5$

The guidelines of Curve Sketching

A. Domain.

We know that $f(x)$ is a polynomial having a domain of $(-\infty, \infty)$

B. Intercepts.

Solving for $y$-intercpt, when $x = 0$

$y = \left(4 - (0)^2 \right)^5 = 1024$

Solving for $x$-intercept, when $y = 0$

$0 = (4-x^2)^5$

We have, $x = \pm 2$

C. Symmetry.

Since $f(-x) = f(x)$, the function is symmetric to $y$-axis.

D. Asymptotes.

None, the function has no denominator

E. Intervals of Increase or Decrease.

If we take the derivative of $f(x)$. By using Chain Rule,

$ \begin{equation} \begin{aligned} f'(x) &= 5(4-x^2)^4(-2x)\\ \\ f'(x) &= -10x(4-x^2)^4 \end{aligned} \end{equation} $

When $f'(x) = 0$, $0 = -10x (4-x^2)^4$

We have, $x =0$ and $-x^2 + 4 = 0$

The critical numbers are, $x = 0$ and $x = \pm 2$

So, the intervals of increase or decrease are...

$ \begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f\\ \hline\\ x < -2 & + & \text{increasing on } (-\infty,-2)\\ \hline\\ -2 < x < 0 & + & \text{increasing on } (-2,0)\\ \hline\\ 0 < x < 2& - & \text{decreasing on } (0,2)\\ \hline\\ x > 2 & - & \text{decreasing on } (2, \infty)\\ \hline \end{array} $

F. Local Maximum and Minimum Values.

Since $f'(x)$ changes from positive to negative at $x=0$, $f(0) = 1024$ is a local maximum. However, since $f'(x)$ does not change from negative to positive, it means that the function has no local minimum.

G. Concavity and Points of Inflection.

$ \begin{equation} \begin{aligned} \text{if } f'(x) &= -10x(4-x^2)^4, \text{ then, by using Product Rule and Chain Rule }\\ \\ f''(x) &= -10 x \cdot 4(4-x^2)^3 (-2x) + (-10) (4-x^2)^4\\ \\ f''(x) &= -10(4-x^2)^3 \left[ -8x^2 + (4-x^2) \right]\\ \\ f''(x) &= -10(4-x^2)^3 \left[ -9x^2 + 4 \right] \end{aligned} \end{equation} $

When $f''(x) =0$

$0 = -10(4-x^2)^3 \left[ -9x^2 + 4 \right]$

We have, $(4-x^2)^3 = 0$ and $-9x^2 + 4 = 0$

The inflection poitns are, $x = \pm 2$ and $\displaystyle x = \pm \frac{2}{3}$

Thus, the concavity can be determined by dividing the interval to...

$ \begin{array}{|c|c|c|} \hline\\ \text{Interval} & f''(x) & \text{Concavity}\\ \hline\\ x < -2 & - & \text{Downward}\\ \hline\\ -2 < x < \frac{-2}{3} & + & \text{Upward}\\ \hline\\ \frac{-2}{3} < x < \frac{2}{3} & - & \text{Downward}\\ \hline\\ \frac{2}{3} < x < 2 & + & \text{Upward}\\ \hline\\ x > 2 & - & \text{Downward}\\ \hline \end{array} $

H. Sketch the Graph.

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