Question

(a) Determine the intervals on which the function \(f(x) = 2{x^3} + 3{x^2} - 36x\) is increasing or decreasing.

(b) Determine the local maximum and minimum values of \(f(x) = 2{x^3} + 3{x^2} - 36x\).

(c) Determine the intervals of concavity and the inflection points of\(f(x) = 2{x^3} + 3{x^2} - 36x\).

Short answer

(a) The function \(f(x)\) is increasing on \(( - \infty , - 3) \cup (2,\infty )\) and decreasing on \(( - 3,2)\).

(b) The local maximum is \(f( - 3) = 81\) and local minimum is \(f(2) = - 44\).

(c) The function \(f(x)\) is concave upward on \(\left( { - \frac{1}{2},\;\infty } \right)\) and \(f(x)\) is concave downward on \(\left( { - \infty ,\; - \frac{1}{2}} \right)\) and the inflection point is \(\left( { - \frac{1}{2},\;\frac{{37}}{2}} \right)\).

Step-by-step solution

Given data

The given function is, \(f(x) = 2{x^3} + 3{x^2} - 36x\).

Concept of Increasing/decreasing 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.

Definition used:

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

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

Obtain the critical points 

Obtain the first derivative of given function as follows:

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {2{x^3} + 3{x^2} - 36x} \right)\\ &= 6{x^2} + 6x - 36\end{aligned}\)

Set \({f^\prime }(x) = 0\) and obtain the critical points as follows:

\(\begin{aligned}{c}6{x^2} + 6x - 36 &= 0\\6\left( {{x^2} + x - 6} \right) &= 0\\6(x - 2)(x + 3) &= 0\\x &= 2,\;x &= - 3\end{aligned}\)

Hence the critical points occurs at \(x = 2,x = - 3\) and in the interval notation\(( - \infty , - 3) \cup ( - 3,2) \cup (2,\infty )\).

Determine the sign of \({f^\prime }(x)\)

Determine the sign of \({f^\prime }(x)\) in table 1 as follows:

Table 1

From the above table observe that, \(f(x)\) is increasing on \(( - \infty , - 3) \cup (2,\infty )\) and decreasing on \(( - 3,\;2)\).

Determine the local maximum and local minimum

(b)

From part (a), the function \(f(x)\) is decreasing on \(( - 3,2)\).

Thus, the local maximum occurs at \(x = - 3\) and the local minimum occurs at \(x = 2\).

Substitute \(x = - 3\) in \(f(x)\) and obtain the local maximum as follows:

\(\begin{aligned}{c}f( - 3) &= 2{( - 3)^3} + 3{( - 3)^2} - 36( - 3)\\ &= 2( - 27) + 3(9) + 108\\ &= - 54 + 27 + 108\\ &= 81\end{aligned}\)

Substitute \(x = 2\) in \(f(x)\) and obtain the local minimum as follows:

\(\begin{aligned}{c}f(2) &= 2{(2)^3} + 3{(2)^2} - 36(2)\\ &= 16 + 12 - 72\\ &= - 44\end{aligned}\)

Thus, the local maximum is \(f( - 3) = 81\) and local minimum is \(f(2) = - 44\).

Determine the interval notation

(c)

Obtain the derivative of \({f^\prime }(x)\) as follows:

\(\begin{aligned}{c}{f^{\prime \prime }}(x) &= \frac{d}{{dx}}\left( {{f^\prime }(x)} \right)\\ &= \frac{d}{{dx}}\left( {6{x^2} + 6x - 36} \right)\\ &= 12x + 6\end{aligned}\)

Set \({f^{\prime \prime }}(x) = 0\) and solve for \(x\) as follows:

\(\begin{aligned}{c}12x + 6 &= 0\\12x &= - 6\\x &= - \frac{1}{2}\end{aligned}\)

Thus, in the interval notation \(\left( { - \infty ,\; - \frac{1}{2}} \right) \cup \left( { - \frac{1}{2},\;\infty } \right)\).

Determine the sign of \({f^{\prime \prime }}(x)\)

Determine the sign of \({f^{\prime \prime }}(x)\) in table 2 as follows:

Table 2

Hence, \(f(x)\) is concave upward on \(\left( { - \frac{1}{2},\;\infty } \right)\) and \(f(x)\) is concave downward on \(\left( { - \infty ,\; - \frac{1}{2}} \right)\).

Here the curve changes from concave downward to concave upward at \(x = - \frac{1}{2}\).

Substitute \(x = - \frac{1}{2}\) in \(f(x)\) and find inflection point as follows:

\(\begin{aligned}{c}f(1) &= 2{\left( { - \frac{1}{2}} \right)^3} + 3{\left( { - \frac{1}{2}} \right)^2} - 36\left( { - \frac{1}{2}} \right)\\ &= 2\left( { - \frac{1}{8}} \right) + 3\left( {\frac{1}{4}} \right) + 18\\ &= - \frac{1}{4} + \frac{3}{4} + 18\end{aligned}\)

Solve further:

\(\begin{aligned}{c}f(1) &= \frac{1}{2} + 18\\ &= \frac{{37}}{2}\end{aligned}\)

Thus, the inflection point is \(\left( { - \frac{1}{2},\frac{{37}}{2}} \right)\).