Question

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

\(\begin{aligned}{l}f'\left( x \right) > 0\;\;{\rm{if}}\;\;\left| x \right| < 2,\;\;\;\;f'\left( x \right) < 0\;\;{\rm{if}}\;\;\left| x \right| > 2,\\f'\left( 2 \right) = 0,\;\;\;\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 1,\;\;f\left( { - x} \right) = - f\left( x \right)\\f''\left( x \right) < 0\;{\rm{if}}\,0 < x < 3,\;\;f''\left( x \right) > 0\;\;{\rm{if}}\;\;x > 3\end{aligned}\)

Short answer

It is given that, \({f^\prime }(x) > 0\) if \(|x| < 2,{f^\prime }(x) < 0\) if \(|x| > 2\) and \({f^\prime }(2) = 0,\mathop {\lim }\limits_{x \to \infty } (x) = 1\) and \(f( - x) = - f(x)\) and also it is given that, \({f^{\prime \prime }}(x) < 0\) if \(0 < x < 3,{f^{\prime \prime }}(x) > 0\) if \(x > 3\).

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 \({f^\prime }(2) = 0,\mathop {\lim }\limits_{x \to \infty } (x) = 1\) and \(f( - x) = - f(x)\) and also it is given that, \({f^{\prime \prime }}(x) < 0\) if \(0 < x < 3,{f^{\prime \prime }}(x) > 0\) if \(x > 3\).

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 \( - 2 < x < 2\) 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 \(x = 2\).

Since \(f( - x) = - f(x)\), the function is an odd function.

Note that, the function would be concave downward in the intervals \((0,3)\) as \({f^{\prime \prime }}(x) < 0\) for \(0 < x < 3\) and concave upward in the intervals \((3,\infty )\) as \({f^{\prime \prime }}(x) > 0\) for \(x > 3\).

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

From the Figure, the horizontal tangent is at \(x = 2\).