Question b2764

Mar 4, 2017

(c) 2 and 3 only

Explanation:

Given:

$f \left(x\right) = {x}^{- \frac{1}{3}}$

Note that:

$f \left(0\right) = \frac{1}{0} \text{ }$ is undefined

${\lim}_{x \to 0 +} {x}^{- \frac{1}{3}} = {\lim}_{t \to 0 +} \frac{1}{t} = \infty$

${\lim}_{x \to 0 -} {x}^{- \frac{1}{3}} = {\lim}_{t \to 0 -} \frac{1}{t} = - \infty$

So 1. is false and 2. is true.

In order to check 3., we need to split the integral into two parts, noting that:

$f \left(x\right) < 0 \text{ }$ when $x \in \left[- 1 , 0\right)$

$f \left(x\right) > 0 \text{ }$ when $x \in \left[1 , 0\right)$

So:

$A = \left\mid {\int}_{- 1}^{0} {x}^{- \frac{1}{3}} \mathrm{dx} \right\mid + \left\mid {\int}_{0}^{1} {x}^{- \frac{1}{3}} \mathrm{dx} \right\mid$

color(white)(A) = abs([3/2 x^(2/3)]_(-1)^0) + abs([3/2 x^(2/3)]_0^1])#

$\textcolor{w h i t e}{A} = \left\mid 0 - \frac{3}{2} \right\mid + \left\mid \frac{3}{2} - 0 \right\mid$

$\textcolor{w h i t e}{A} = \frac{3}{2} + \frac{3}{2}$

$\textcolor{w h i t e}{A} = 3$

So the area is finite and 3. is true.

graph{(y-x^(-1/3))(x+1+0.00001y)(x-1+0.00001y) = 0 [-5.086, 4.92, -2.55, 2.45]}