Answer to Question #223131 in Calculus for Simen

Question

Study the following functions

3) f(x) = (x³ / x-1 )s

4) f(x) = x / cube root of (x²-1)

Answer 1

Number 3

$\mathrm{Domain\:of\:}\:\frac{x^3}{x-1}\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x<1\quad \mathrm{or}\quad \:x>1\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:1\right)\cup \left(1,\:\infty \:\right)\end{bmatrix}\\ \mathrm{Range\:of\:}\frac{x^3}{x-1}:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<f\left(x\right)<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}\\ \mathrm{Axis\:interception\:points\:of}\:\frac{x^3}{x-1}:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)\\ \mathrm{Asymptotes\:of}\:\frac{x^3}{x-1}:\quad \mathrm{Vertical}:\:x=1\\ \mathrm{Extreme\:Points\:of}\:\frac{x^3}{x-1}:\quad \mathrm{Saddle}\left(0,\:0\right),\:\mathrm{Minimum}\left(\frac{3}{2},\:\frac{27}{4}\right)$

Number 4

$\mathrm{Domain\:of\:}\:\frac{x}{\sqrt[3]{x^2-1}}\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x<-1\quad \mathrm{or}\quad \:-1<x<1\quad \mathrm{or}\quad \:x>1\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:-1\right)\cup \left(-1,\:1\right)\cup \left(1,\:\infty \:\right)\end{bmatrix}\\ \mathrm{Range\:of\:}\frac{x}{\sqrt[3]{x^2-1}}:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)>\sqrt[3]{-1}\cdot \infty \\ \:\mathrm{Interval\:Notation:}&\:\left(\sqrt[3]{-1}\cdot \infty \:,\:\infty \:\right)\end{bmatrix}\\ \mathrm{Axis\:interception\:points\:of}\:\frac{x}{\sqrt[3]{x^2-1}}:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)\\ \mathrm{Asymptotes\:of}\:\frac{x}{\sqrt[3]{x^2-1}}:\quad \mathrm{Vertical}:\:x=-1,\:x=1\\ \mathrm{Extreme\:Points\:of}\:\frac{x}{\sqrt[3]{x^2-1}}:\quad \mathrm{Maximum}\left(-\sqrt{3},\:-\frac{\sqrt{3}}{\sqrt[3]{2}}\right),\:\mathrm{Minimum}\left(\sqrt{3},\:\frac{\sqrt{3}}{\sqrt[3]{2}}\right)$