Question

hw 16 - first derivative test section 3.2: problem 4
(1 point)
consider the function
$$ f ( x ) = x ^ { 3 } + 9 x ^ { 2 } - 12. $$
a) find the critical numbers of ( f ). (separate multiple answers by commas.)
b) determine the intervals on which ( f ) is increasing and decreasing.
( f ) is increasing on:
( f ) is decreasing on:

c) use the first derivative test to determine whether each critical point is a relative maximum, minimum, or neither.
relative maxima occur at ( x = ) (separate multiple answers by commas.)
relative minima occur at ( x = ) (separate multiple answers by commas.)

Explanation:

Step1: Find the derivative of \(f(x)\)

The derivative of \(f(x)=x^{3}+9x^{2}-12\) using the power rule \((x^{n})^\prime = nx^{n - 1}\) is \(f^\prime(x)=3x^{2}+18x=3x(x + 6)\)

Step2: Find the critical numbers

Set \(f^\prime(x)=0\), so \(3x(x + 6)=0\).
Solving \(3x(x + 6)=0\) gives \(x = 0\) or \(x=-6\) (critical numbers).

Step3: Determine the intervals of increase and decrease

We consider the intervals \((-\infty,-6)\), \((-6,0)\) and \((0,\infty)\)

• For \(x\in(-\infty,-6)\), let \(x=-7\), then \(f^\prime(-7)=3\times(-7)\times(-7 + 6)=21>0\)
• For \(x\in(-6,0)\), let \(x=-1\), then \(f^\prime(-1)=3\times(-1)\times(-1 + 6)=-15<0\)
• For \(x\in(0,\infty)\), let \(x = 1\), then \(f^\prime(1)=3\times1\times(1 + 6)=21>0\)

So \(f(x)\) is increasing on \((-\infty,-6)\cup(0,\infty)\) and decreasing on \((-6,0)\)

Step4: Use the First - Derivative Test

• At \(x=-6\): \(f^\prime(x)\) changes from positive (left of \(x = - 6\)) to negative (right of \(x=-6\)), so \(x=-6\) is a relative maximum.
• At \(x = 0\): \(f^\prime(x)\) changes from negative (left of \(x = 0\)) to positive (right of \(x = 0\)), so \(x = 0\) is a relative minimum.

Answer:

a) \(-6,0\)
b) \(f\) is increasing on: \((-\infty,-6),(0,\infty)\); \(f\) is decreasing on: \((-6,0)\)
c) Relative maxima occur at \(x=-6\); Relative minima occur at \(x = 0\)