Question

f(x)=\frac{x^{4}}{4}-4x^{3}-2.
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.)
c) find the critical numbers of f and use the second derivative test, when possible, to determine the relative extrema. list only the x - coordinates.
relative maxima at: (separate multiple answers by commas.)
relative minima at: (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

The function is \(f(x)=\frac{x^{4}}{4}-4x^{3}-2\).
The first derivative \(f^{\prime}(x)=x^{3}-12x^{2}\) (using the power rule \((x^{n})^\prime = nx^{n - 1}\)).
The second derivative \(f^{\prime\prime}(x)=3x^{2}-24x=3x(x - 8)\).

Step2: Find the intervals of concavity

Set \(f^{\prime\prime}(x)=0\), so \(3x(x - 8)=0\), which gives \(x = 0\) and \(x = 8\).
Test intervals:

• For \(x<0\), let \(x=-1\), then \(f^{\prime\prime}(-1)=3\times(-1)\times(-1 - 8)=27>0\).
• For \(0
• For \(x>8\), let \(x = 9\), then \(f^{\prime\prime}(9)=3\times9\times(9 - 8)=27>0\).

So \(f(x)\) is concave up on \((-\infty,0)\cup(8,\infty)\) and concave down on \((0,8)\).

Step3: Find inflection points

Since the concavity changes at \(x = 0\) and \(x = 8\).
When \(x = 0\), \(y=f(0)=\frac{0^{4}}{4}-4\times0^{3}-2=-2\).
When \(x = 8\), \(y=f(8)=\frac{8^{4}}{4}-4\times8^{3}-2=\frac{4096}{4}-4\times512-2=1024 - 2048-2=-1026\).
The inflection points are \((0,-2),(8,-1026)\).

Step4: Find critical numbers and relative extrema

Set \(f^{\prime}(x)=0\), \(x^{3}-12x^{2}=x^{2}(x - 12)=0\), so \(x = 0\) and \(x = 12\).
Now use the second - derivative test:

• For \(x = 0\), \(f^{\prime\prime}(0)=3\times0\times(0 - 8)=0\), the second - derivative test fails.
• For \(x = 12\), \(f^{\prime\prime}(12)=3\times12\times(12 - 8)=144>0\). So \(x = 12\) is a relative minimum.

Answer:

a) \(f\) is concave up on \((-\infty,0)\cup(8,\infty)\), \(f\) is concave down on \((0,8)\).
b) \((0,-2),(8,-1026)\)
c) Relative maxima: DNE, Relative minima: \(12\)