Question

To sketch the graph of a function that satisfies the given condition.

\(\begin{aligned}{l}f'\left( x \right) > 0\;\;{\rm{if}}\;\;\left| x \right| < 2,\;\;\;\;f'\left( x \right) < 0\;\;{\rm{if}}\;\;\left| x \right| > 2,\\f'\left( { - 2} \right) = 0,\;\;\;\mathop {\lim }\limits_{x \to 2} \left| {f'\left( x \right)} \right| = \infty ,\;\;f''\left( x \right) > 0\;{\rm{if}}\,x \ne 2\end{aligned}\)

Short answer

It is given that, \({f^\prime }(x) > 0\) if \(|x| < 2,{f^\prime }(x) < 0\) if \(|x| > 2\) and \({f^\prime }( - 2) = 0\), \(\mathop {\lim }\limits_{x \to 2} \left| {{f^\prime }(x)} \right| = \infty \) and also it is given that, \({f^{\prime \prime }}(x) > 0\) if \(x \ne 2\).

Step-by-step solution

Given data

It is given that, \({f^\prime }(x) > 0\) if \(|x| < 2,{f^\prime }(x) < 0\) if \(|x| > 2\) and \({f^\prime }( - 2) = 0\), \(\mathop {\lim }\limits_{x \to 2} \left| {{f^\prime }(x)} \right| = \infty \) and also it is given that, \({f^{\prime \prime }}(x) > 0\) if \(x \ne 2\).

Concept of Increasing test / decreasing test and concavity test

Increasing/Decreasing Test:

(a) If\({f^\prime }(x) > 0\)on an interval, then\(f\)is increasing on that interval.

(b) If\({f^\prime }(x) < 0\)on an interval, then\(f\)is decreasing on that interval".

Concavity Test:

If\({f^{\prime \prime }}(x) > 0\)for all\(x\)in\(I\), then the graph of\(f\)is concave upward on\({f^{\prime \prime }}(x) < 0\)for all\(x\)in\(I\), then the graph of\(f\)is concave downward on\({I^{\prime \prime }}\).

Sketch of the graph which satisfy the given conditions

From the increasing test, as \({f^\prime }(x) > 0\) for \(|x| < 2\) the function is increasing on the interval \(( - 2,2)\).

Similarly, from the decreasing test, as \({f^\prime }(x) < 0\) for \(|x| > 2\) the interval in which the function is decreasing is \(( - \infty , - 2) \cup (2,\infty )\).

Since the first derivative at \( - 2\) is 0 , the horizontal tangent is at \(x = - 2\).

Also note that, \(\mathop {\lim }\limits_{x \to 2} \left| {{f^\prime }(x)} \right| = \infty \) then there is a vertical asymptote at \(x = 2\).

Note that, the function would be concave upward in the intervals \(( - \infty ,2) \cup (2,\infty )\) as \({f^{\prime \prime }}(x) > 0\) for \(x \ne 2\).

Sketch the graph of the given function using the given information as:

From the Figure, it is observed the vertical asymptote is \(x = 2\).