Question

Find the intervals on which\(f\)is increasing or decreasing, and find the local maximum and minimum values of\(f\).

10. \(f\left( x \right) = {x^3} - 6{x^2} - 135x\)

Short answer

The function \(f\) changes from increasing to decreasing at \(x = - 5\) and from decreasing to increasing at \(x = 9\).

The local maximum value is \(f\left( { - 5} \right) = 400\) and the local minimum value is \(f\left( 9 \right) = - 972\).

Step-by-step solution

Write the formula increasing and decreasing test

If the function \(f'\left( x \right) > 0\) on an interval, then the function \(f\) is said to be increasing on that interval.

If the function \(f'\left( x \right) < 0\) on an interval, then the function \(f\) is said to be decreasing on that interval.

Step 2: Find the solution of the function

Consider the function \(f\left( x \right) = {x^3} - 6{x^2} - 135x\).

Differentiate the function w.r.t.\(x\).

\(\begin{array}{c}\frac{d}{{dx}}f\left( x \right) = \frac{d}{{dx}}\left( {{x^3} - 6{x^2} - 135x} \right)\\ = \frac{d}{{dx}}\left( {{x^3}} \right) - 6\frac{d}{{dx}}\left( {{x^2}} \right) - 135\frac{d}{{dx}}\left( x \right)\\ = 3{x^2} - 12x - 135\\ = 3\left( {{x^2} - 4x - 45} \right)\\ = 3\left( {x + 5} \right)\left( {x - 9} \right)\end{array}\)

Since \(f'\left( x \right) = 0\) then find the solutions of the functions.

\(\begin{array}{c}3\left( {x + 5} \right)\left( {x - 9} \right) = 0\\x = - 5,\;9\end{array}\)

Step 3: construct the table to check whether the function is decreasing or increasing

Now construct the table as shown below:

Interval	\(x + 5\)	\(x - 9\)	\(f'\left( x \right)\)	\(f\)
\(x < - 5\)	\( - \)	\( - \)	\( + \)	Increasing on \(\left( { - \infty , - 5} \right)\)
\( - 5 < x < 9\)	\( + \)	\( - \)	\( - \)	decreasing on \(\left( { - 5,9} \right)\)
\(x > 9\)	\( + \)	\( + \)	\( + \)	decreasing on \(\left( {9,\infty } \right)\)

On observing the table \(f\) changes from increasing to decreasing at \(x = - 5\) and from decreasing to increasing at \(x = 9\) .

Step 4: Find local maxima and minima

Put\(x = - 5\)and \(x = 9\) into \(f\left( x \right) = {x^3} - 6{x^2} - 135x\).

\(\begin{array}{c}f\left( { - 5} \right) = {\left( { - 5} \right)^3} - 6{\left( { - 5} \right)^2} - 135\left( { - 5} \right)\\ = - 125 - 150 + 675\\ = 400\end{array}\)

And

\(\begin{array}{c}f\left( 9 \right) = {\left( 9 \right)^3} - 6{\left( 9 \right)^2} - 135\left( 9 \right)\\ = 729 - 486 - 1215\\ = - 972\end{array}\)

Thus, the local maximum value is \(f\left( { - 5} \right) = 400\) and the local minimum value is \(f\left( 9 \right) = - 972\).