Question

55-58 The graph of a function f is shown. (The dashed lines indicate horizontal asymptotes). Find each of the following for the given function g.

a) The domain of g and \(g'\)

b) The critical numbers of g

c) The approximate value of \(g'\left( {\bf{6}} \right)\)

d) All vertical and horizontal asymptotes of g

56. \(g\left( x \right) = \sqrt[{\bf{3}}]{{f\left( x \right)}}\)

Short answer

a) The domain of g is \(\left( { - \infty ,\infty } \right)\) , and domain of \(g'\) is \(\left( { - \infty ,3} \right) \cup \left( {3,7} \right) \cup \left( {7,\infty } \right)\).

b) The critical numbers of g are 3, 5, 7, and 9.

c) \( - 0.32\)

d) The curve of g has horizontal asymptote \(y = \sqrt[3]{2}\) and \(y = 1\). There is no vertical asymptote.

Step-by-step solution

Find an answer for part (a)

As the cube root of any function is defined for all real numbers, so the domain of g is\(\left( { - \infty ,\infty } \right)\).

Find the derivative of the function\(g\left( x \right)\).

\(\begin{array}{c}g'\left( x \right) = \frac{1}{{3\left( {\sqrt[3]{{f\left( x \right)}}} \right)}} \cdot f'\left( x \right)\\ = \frac{{f'\left( x \right)}}{{3\left( {\sqrt[3]{{f\left( x \right)}}} \right)}}\end{array}\)

As\(f'\left( 3 \right)\)does not exist and\(f\left( 7 \right) = 0\), therefore\(g'\left( 3 \right)\)and\(g\left( 7 \right)\)do not exist.

The domain of \(g'\) is \(\left( { - \infty ,3} \right) \cup \left( {3,7} \right) \cup \left( {7,\infty } \right)\).

Find an answer for part (b)

If\(g'\left( x \right) = 0\), then\(f'\left( x \right) = 0\).

From the figure, it can be observed that at\(x = 5\)and\(x = 9\)are horizontal tangent lines. So,\(g'\left( x \right)\)does not exist at\(x = 3\)and\(x = 7\).

Thus, the critical numbers are 3, 5, 7, and 9.

Find an answer for part (c)

Substitute 6 for x in the equation \(g'\left( x \right) = \frac{{f'\left( x \right)}}{{3\left( {\sqrt[3]{{f\left( x \right)}}} \right)}}\).

\(\begin{array}{c}g'\left( 6 \right) = \frac{{f'\left( 6 \right)}}{{3\left( {\sqrt[3]{{f\left( 6 \right)}}} \right)}}\\ = \frac{{ - 2}}{{3{{\left( {\sqrt[3]{3}} \right)}^2}}}\\ = - \frac{2}{{{3^{\frac{5}{3}}}}}\\ \approx - 0.32\end{array}\)

So, the value of \(g'\left( 6 \right)\) is \( - 0.32\).

Find an answer for part (d)

It can be observed from the graph that there is no vertical asymptote.

The horizontal asymptotes are shown below:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to - \infty } g\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \sqrt[3]{{f\left( x \right)}}\\ = \sqrt[3]{2}\end{array}\)

And,

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } g\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \sqrt[3]{{f\left( x \right)}}\\ = \mathop {\lim }\limits_{x \to \infty } \sqrt[3]{{\left( { - 1} \right)}}\\ = - 1\end{array}\)

So, the horizontal asymptotes are \(y = \sqrt[3]{2}\) and \(y = 1\).