Chapter 12, Problem 6PS

a.

To determine

To find: The domain of $f$ .

Answer to Problem 6PS

Domain of $f$ is $[-27,1)\cup (1,\infty )$ .

Explanation of Solution

Given Information: Given function is $f(x)=\frac{\sqrt{3+x^{1/3}}-2}{x-1}$ .

For this function to be defined,

  $\begin{array}{l}3+x^{1/3}\ge 0\\ \Rightarrow x^{1/3}\ge -3\\ $ \Rightarrow x\ge {(-3)}^3$\Rightarrow x\ge -27\end{array}$

and $\begin{array}{l}x-1\ne 0\\ \Rightarrow x\ne 1\end{array}$

So domain of $f$ is $[-27,1)\cup (1,\infty )$ .

b.

To determine

To graph: The given function.

Explanation of Solution

Given Information: Given function is $f(x)=\frac{\sqrt{3+x^{1/3}}-2}{x-1}$ .

Graph:

  

c.

To determine

To evaluate: $\underset{x\to -{27}^{+}}{\lim }f(x)$ and to verify it using the graph in part (b).

Answer to Problem 6PS

  $\underset{x\to -{27}^{+}}{\lim }f(x)=\mathrm{0.07143.}$

Explanation of Solution

Given Information: Given function is $f(x)=\frac{\sqrt{3+x^{1/3}}-2}{x-1}$ .

Calculation:

  $\underset{x\to -{27}^{+}}{\lim }f(x)=\frac{\sqrt{3+{(-27)}^{1/3}}-2}{-27-1}=\frac{\sqrt{3-3}-2}{-28}=\frac{0-2}{-28}=\frac{1}{14}=\mathrm{0.07143.}$

Thus, $\underset{x\to -{27}^{+}}{\lim }f(x)=\mathrm{0.07143.}$

  

From the graph, as $x$ approaches $-27$ from the right hand side, $f(x)$ approaches $0.07143$ .

Hence, verified.

Thus, $\underset{x\to -{27}^{+}}{\lim }f(x)=\mathrm{0.07143.}$

d.

To determine

To evaluate: $\underset{x\to 1}{\lim }f(x)$ and to verify it using the graph in part (b).

Answer to Problem 6PS

  $\underset{x\to 1}{\lim }f(x)=\frac{1}{12}$

Explanation of Solution

Given Information: Given function is $f(x)=\frac{\sqrt{3+x^{1/3}}-2}{x-1}$ .

Calculation:

Let $x^{1/3}=z$ , i.e., $x=z^3$

As $x\to 1,z\to 1$ .

So $f(z)=\frac{\sqrt{3+z}-2}{z^3-1}$

  $\Rightarrow \underset{z\to 1}{\lim }f(z)=\underset{z\to 1}{\lim }\frac{\sqrt{3+z}-2}{z^3-1}$

  $\begin{array}{l}=\underset{z\to 1}{\lim }\frac{(\sqrt{3+z}-2)(\sqrt{3+z}+2)}{(z^3-1)(\sqrt{3+z}+2)}\\ =\underset{z\to 1}{\lim }\frac{3+z-4}{(z-1)(z^2+z+1)(\sqrt{3+z}+2)}\\ =\underset{z\to 1}{\lim }\frac{z-1}{(z-1)(z^2+z+1)(\sqrt{3+z}+2)}\\ =\underset{z\to 1}{\lim }\frac{1}{(z^2+z+1)(\sqrt{3+z}+2)}\\ =\frac{1}{(1^2+1+1)(\sqrt{3+1}+2)}\\ =\frac{1}{3(4)}\\ =\frac{1}{12}\end{array}$

  $=0.0833$

  

From the graph, as $x$ approaches $1$ , $f(x)$ approaches $0.0833$ .

Hence verified.

Thus, $\underset{x\to 1}{\lim }f(x)=\frac{1}{12}$