## Calculus: Early Transcendentals 8th Edition

A. The domain is $(-\infty, \infty)$ B. The y-intercept is $1$ The x-intercept is $-1$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} \sqrt[3] {x^3+1} = -\infty$ $\lim\limits_{x \to \infty} \sqrt[3] {x^3+1} = \infty$ The line $y=x$ is a slant asymptote. E. The function is increasing on the interval $(-\infty, \infty)$ F. There is no local minimum or local maximum. G. The graph is concave down on the interval $(-1, 0)$ The graph is concave up on the intervals $(-\infty,-1)\cup (0, \infty)$ The points of inflection are $(-1,0)$ and $(0,1)$ H. We can see a sketch of the curve below.
$y = \sqrt[3] {x^3+1}$ A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$ B. When $x=0$, then $y = \sqrt[3] {0^3+1} = 1$ The y-intercept is $1$ When $y = 0$: $\sqrt[3] {x^3+1} = 0$ $x^3+1 = 0$ $x =-1$ The x-intercept is $-1$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} \sqrt[3] {x^3+1} = -\infty$ $\lim\limits_{x \to \infty} \sqrt[3] {x^3+1} = \infty$ There are no horizontal asymptotes. $\lim\limits_{x \to -\infty} (\sqrt[3] {x^3+1}-x) = 0$ $\lim\limits_{x \to \infty} (\sqrt[3] {x^3+1}-x) = 0$ The line $y=x$ is a slant asymptote. E. We can find values of $x$ such that $y' = 0$: $y' =\frac{1}{3}(x^3+1)^{-2/3}(3x^2) = \frac{x^2}{(x^3+1)^{2/3}} = 0$ $x^2 = 0$ $x = 0$ Note that $y'$ is undefined when $x=-1$ When $x \lt -1$ or $-1 \lt x \lt 0$ or $x \gt 0$, then $y' \gt 0$ The function is increasing on the interval $(-\infty, \infty)$ F. There is no local minimum or local maximum since the function is increasing on all intervals. G. We can try to find the values of $x$ such that $y'' = 0$: $y'' =\frac{(2x)((x^3+1)^{2/3})- (x^2)(\frac{2x^2}{(x^3+1)^{1/3}})}{(x^3+1)^{4/3}}$ $y'' =\frac{(2x)(x^3+1)- (x^2)(2x^2)}{(x^3+1)^{5/3}}$ $y'' =\frac{2x}{(x^3+1)^{5/3}} = 0$ $2x = 0$ $x = 0$ Note that $y''$ is undefined when $x=-1$ When $-1 \lt x \lt 0$, then $y'' \lt 0$ The graph is concave down on the interval $(-1, 0)$ When $x \lt -1$ or $x \gt 0$, then $y'' \gt 0$ The graph is concave up on the intervals $(-\infty,-1)\cup (0, \infty)$ When $x=-1$, then $y = \sqrt[3] {(-1)^3+1} = 0$ When $x=0$, then $y = \sqrt[3] {0^3+1} = 1$ The points of inflection are $(-1,0)$ and $(0,1)$ H. We can see a sketch of the curve below.