Question

consider the function

f(x)=\frac{x^{4}}{4}-3 x^{3}-3

a) determine the intervals on which ( f ) is concave up and concave down.
( f ) is concave up on:
( f ) is concave down on:

b) based on your answer to part (a), determine the inflection points of ( f ). each point should be entered as an ordered pair (that
is, in the form ( (x, y) ) ).
(separate multiple answers by commas.)

(round to three decimal places at needed.)

note: when using interval notation in webwork, remember that:
you use inf for ( infty ) and -inf for ( -infty ),
and use u for the union symbol.
enter dne if an answer does not exist.

Explanation:

Step1: Find the first and second derivatives

Given \( f(x)=\frac{x^{4}}{4}-3x^{3}-3 \).
The first derivative \( f^{\prime}(x)=x^{3}-9x^{2}\) (using the power rule \((x^{n})^\prime = nx^{n - 1}\)).
The second derivative \( f^{\prime\prime}(x)=3x^{2}-18x=3x(x - 6)\).

Step2: Find the critical points of \( f^{\prime\prime}(x)\)

Set \( f^{\prime\prime}(x)=0\), so \(3x(x - 6)=0\).
Solving \(3x(x - 6)=0\) gives \(x = 0\) and \(x=6\).

Step3: Test intervals for concavity

• For the interval \((-\infty,0)\), let \(x=-1\). Then \(f^{\prime\prime}(-1)=3\times(-1)\times(-1 - 6)=21>0\).
• For the interval \((0,6)\), let \(x = 1\). Then \(f^{\prime\prime}(1)=3\times1\times(1 - 6)=-15<0\).
• For the interval \((6,\infty)\), let \(x = 7\). Then \(f^{\prime\prime}(7)=3\times7\times(7 - 6)=21>0\).

Step4: Find \(y\) - values for inflection points

When \(x = 0\), \(y=f(0)=\frac{0^{4}}{4}-3\times0^{3}-3=-3\).
When \(x = 6\), \(y=f(6)=\frac{6^{4}}{4}-3\times6^{3}-3=\frac{1296}{4}-3\times216-3=324 - 648-3=-327\).

Answer:

a) \(f\) is concave up on: \((-\infty,0)\cup(6,\infty)\)
\(f\) is concave down on: \((0,6)\)
b) \((0,-3),(6,-327)\)