Answer to Question #104424 in Calculus for Akshay

Question

Trace the curve y =³√ x −1 , and state all the properties you use to trace it.

Answer 1

$y=f(x)=\sqrt[3]{x} - 1$

1) Domain

Domain is all real numbers.

2) Symmetry

$f(-x) = \sqrt[3]{-x}-1 = -\sqrt[3]{x}-1 \neq -f(x) \text{ or } f(x)$

The function is neither even nor odd.

3) Intercepts  on the axes

x-intercept:

$y=0 \\ \sqrt[3]{x} - 1 =0\\ x=1$

y-intercept:

$x=0\\ y=\sqrt[3]{0}-1=-1$

4) Asymptotes

Let $y=kx+b$ is asymptote.

$k=\lim\limits_{x\to\infty}{\frac{f(x)}{x}}=\lim\limits_{x\to\infty}{\frac{\sqrt[3]{x}-1}{x}}=0$

$b=\lim\limits_{x\to\infty}{(f(x)-kx)}=\lim\limits_{x\to\infty}{\sqrt[3]{x}-1}=\infty$

No asymptotes.

5) Monotonicity

$y'=\frac{1}{3\sqrt[3]{x^2}}$

The first derivative does not exist when $x=0.$ If $x>0,$ then $y'>0.$ If $x<0,$ then $y'>0.$ Therefore, $y=f(x)$ is monotonically increasing and $x=0$ is the critical point.

6) Convexity

$y'' = -\frac{2}{9x\sqrt[3]{x^2}}$

The second derivative does not exist when $x=0.$ If $x>0,$ then $y''<0.$ If $x<0,$ then $y''>0.$ Therefore, when $x<0$ the function is convex and when $x>0$ the function is concave.