Question

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

\(\begin{aligned}{l}{f^\prime }(1) = {f^\prime }( - 1) = 0,f'\left( x \right) < 0\;{\rm{if}}\;\left| x \right| < 1,\\f'\left( x \right) > 0\;{\rm{if}}\;1 < \left| x \right| < 2,f'\left( x \right) = - 1\;{\rm{if}}\;\left| x \right| > 2\\f''\left( x \right) < 0\;{\rm{if}}\; - 2 < x < 0,\;\;{\rm{inflectionpoint}}\left( {0,1} \right)\end{aligned}\)

Short answer

The graph of a function that satisfies the condition is given as:

Step-by-step solution

Given data

It is given that, \({f^\prime }(1) = {f^\prime }( - 1) = 0\) and \({f^\prime }(x) < 0\) if \(|x| < 1,{f^\prime }(x) > 0\) if \(1 < |x| < 2\) and \({f^\prime }(x) = 1\) if \(|x| > 2\) and also, \({f^{\prime \prime }}(x) < 0\) if \( - 2 < x < 0\) and the inflection point \((0,1)\).

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

It is observed that, \({f^\prime }(1) = {f^\prime }( - 1) = 0\) the local extrema occur at \(x = \pm 1\).

From the increasing test, as \({f^\prime }(x) > 0\) for \(1 < |x| < 2\) the function is increasing on the interval \(( - 2. - 1) \cup (1,2)\).

Similarly, from the decreasing test, as \({f^\prime }(x) < 0\) for \(|x| < 1\) the interval in which the function is decreasing is \(( - 1,1)\).

Note that, the function would be concave upward in the intervals \((0,2)\) as the inflection point is \((0,1)\) and the local minimum is at \(x = 1\).

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

From the Figure, it is observed the inflection point is \((0,1)\).