Increasing, Decreasing Intervals & Extreme Values of f(x) = −x³ + 12x + 5 | Calculus Worked Example 7| Differential Calculus |

Applications of Derivatives · Worked Example 7

Where does f(x) = −x³ + 12x + 5 rise and fall?

Locating monotonic intervals and absolute extrema on a closed domain, −3 ≤ x ≤ 3.

decreasing increasing critical point

Step 1

Find the critical values

Start with the function and differentiate it with respect to \(x\):

\[ \begin{aligned} f(x) &= -x^3 + 12x + 5 \\ f'(x) &= -3x^2 + 12 \end{aligned} \]

Set \(f'(x) = 0\) and solve for \(x\):

\[ \begin{aligned} -3x^2 + 12 &= 0 \\ -3(x^2 - 4) &= 0 \\ x^2 &= 4 \\ x &= -2, \quad x = 2 \end{aligned} \]

Both values lie inside the given interval \([-3, 3]\).

Critical values: \(x = -2,\ x = 2\)

Step 2

Test increasing and decreasing intervals

The two critical points split \([-3, 3]\) into three sub-intervals. Checking the sign of \(f'(x) = -3(x^2-4)\) in each tells us whether \(f\) is rising or falling there.

Sign analysis of f′(x)

Interval	Sign of f′(x)	Behaviour
[−3, −2)	negative (−)	strictly decreasing
(−2, 2)	positive (+)	strictly increasing
(2, 3]	negative (−)	strictly decreasing

Step 3

Evaluate the extreme values

Because the domain is closed, the absolute extrema occur either at a critical point or at an endpoint. We evaluate \(f(x)\) at all four:

Endpoint · x = −3

\( f(-3) = 27 - 36 + 5 = -4 \)

minimum

Critical point · x = −2

\( f(-2) = 8 - 24 + 5 = -11 \)

maximum

Critical point · x = 2

\( f(2) = -8 + 24 + 5 = 21 \)

Endpoint · x = 3

\( f(3) = -27 + 36 + 5 = 14 \)

Comparing all four values, the largest is 21 and the smallest is −11.

Summary

• Increasing on(−2, 2)
• Decreasing on[−3, −2) ∪ (2, 3]
• Maximum value21 at x = 2
• Minimum value−11 at x = −2

Watch it worked out

A step-by-step video walkthrough of this same problem.

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