Question

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 domains of \(g\) and \(g'\)

(b) The critical numbers of \(g\)

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

(d) All vertical and horizontal asymptotes of \(g\).

55. \(g\left( x \right) = \sqrt {f\left( x \right)} \)

Short answer

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

(b) The critical number of \(g\) are 3 and 5.

(c) The approximate value is \(g'\left( 6 \right) \approx - 0.58\).

(d) The line \(y = \sqrt 2 \) is a horizontal asymptote. There is no vertical asymptote.

Step-by-step solution

Find the domain of \(g\) and \(g'\)

(e) The domain of \(g\) consisting of every \(x\) in which \(f\left( x \right) \ge 0\). Therefore, the domain of \(g\) is \(\left( { - \infty ,7} \right)\).

Obtain the derivative of \(g\) as shown below:

\(g'\left( x \right) = \frac{1}{{2\sqrt {f\left( x \right)} }} \cdot f'\left( x \right) = \frac{{f'\left( x \right)}}{{2\sqrt {f\left( x \right)} }}\)

The value of \(g'\left( 3 \right)\) does not exist because \(f'\left( 3 \right)\) does not exist. It is observed that \(f\left( 7 \right) = 0\), however, the endpoint in the domain of \(g\) is 7.

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

Determine the critical number of g

(f)

When \(g'\left( x \right) = 0\) then the domain of \(g\) has \(f'\left( x \right) = 0\). According to part (a), \(g'\left( 3 \right)\) does not exist. Therefore, the critical number of \(g\) are 3 and 5.

Determine the approximate value of \(g'\left( 6 \right)\)

(c)

According to part (a), that the derivative of \(g\) is \(g'\left( x \right) = \frac{{f'\left( x \right)}}{{2\sqrt {f\left( x \right)} }}\).

Obtain the value of \(g'\left( 6 \right)\) as shown below:

\(\begin{array}{c}g'\left( 6 \right) = \frac{{f'\left( 6 \right)}}{{2\sqrt {f\left( 6 \right)} }}\\ = \frac{{ - 2}}{{2\sqrt {f\left( 6 \right)} }}\\ \approx \frac{{ - 2}}{{2\sqrt 3 }}\\ = - \frac{1}{{\sqrt 3 }}\\ \approx - 0.58\end{array}\)

Thus, the approximate value is \(g'\left( 6 \right) \approx - 0.58\).

Determine the vertical and horizontal asymptotes

(d)

Obtain the horizontal asymptote as shown below:

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

Thus, the line \(y = \sqrt 2 \) is a horizontal asymptote. There is no vertical asymptote.