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. \(g\left( x \right) = \frac{1}{{f\left( x \right)}}\)

Short answer

a) The domain of g is \(\left( { - \infty ,7} \right) \cup \left( {7,\infty } \right)\) , 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) \(\frac{2}{9}\)

d) The curve of g has horizontal asymptote \(y = \frac{1}{2}\) and \(y = - 1\). And \(x = 7\) is a vertical asymptote.

Step-by-step solution

Find an answer for part (a)

The function g is consist at all points except where\(f\left( x \right) \ne 0\). Then domain of g is\(\left( { - \infty ,7} \right) \cup \left( {7,\infty } \right)\).

The derivative of\(g\left( x \right)\)can be expressed as:

\(\begin{array}{c}g'\left( x \right) = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{1}{{f\left( x \right)}}} \right)\\ = - \frac{{f'\left( x \right)}}{{{{\left( {f\left( x \right)} \right)}^2}}}\end{array}\)

As \(f'\left( 3 \right)\) does not exist and \(f\left( 7 \right) = 0\). 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\(x = 9\)there are horizontal tangent lines.\(g'\left( x \right)\)does not exist at\(x = 3\)and\(x = 7\).

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

Find an answer for part (c)

Substitute 6 for x in the equation \(g'\left( x \right) = - \frac{{f'\left( x \right)}}{{{{\left( {f\left( x \right)} \right)}^2}}}\).

\(\begin{array}{c}g'\left( 6 \right) = - \frac{{f'\left( 6 \right)}}{{{{\left( {f\left( 6 \right)} \right)}^2}}}\\ = - \frac{{\left( { - 2} \right)}}{{{3^2}}}\\ = \frac{2}{9}\end{array}\)

So, the value of \(g'\left( 6 \right)\) is \(\frac{2}{9}\).

Find an answer for part (d)

For horizontal asymptote:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to - \infty } g\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{{f\left( x \right)}}\\ = \frac{1}{2}\end{array}\)

And,

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

So, the horizontal asymptotes are \(y = \frac{1}{2}\) and \(y = - 1\).

For vertical asymptote:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {7^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {7^ - }} \frac{1}{{f\left( x \right)}}\\ = \infty \end{array}\)

And,

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {7^ + }} g\left( x \right) = \mathop {\lim }\limits_{x \to {7^ + }} \frac{1}{{f\left( x \right)}}\\ = - \infty \end{array}\)

So, \(x = 7\) is a vertical asymptote.