Question

find the first and second derivatives of the function f(x)=x^4 - 24x^2 + 64x + 16 and report the x - coordinates of any relative extrema and inflection points.
a. find formulas for f(x) and f(x). the graphs of f(x), f(x), and f(x) are shown to the right.
f(x)=
f(x)=

Explanation:

Step1: Find the first - derivative

Use the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$.
For $f(x)=x^{4}-24x^{2}+64x + 16$, we have $f'(x)=4x^{3}-48x + 64$.

Step2: Find the second - derivative

Differentiate $f'(x)$ with the power rule.
$f''(x)=\frac{d}{dx}(4x^{3}-48x + 64)=12x^{2}-48$.

Step3: Find relative extrema

Set $f'(x)=0$, so $4x^{3}-48x + 64 = 0$. Divide through by 4: $x^{3}-12x + 16=0$.
By trial - and - error or factoring, we find that $(x - 2)(x^{2}+2x - 8)=0$.
Further factoring gives $(x - 2)(x - 2)(x+4)=0$. So $x=-4$ and $x = 2$ are the critical points.

Step4: Classify relative extrema

Use the second - derivative test.
When $x=-4$, $f''(-4)=12\times(-4)^{2}-48=192 - 48 = 144>0$. So $f(x)$ has a relative minimum at $x=-4$.
When $x = 2$, $f''(2)=12\times2^{2}-48=48 - 48 = 0$. The second - derivative test is inconclusive. We can use the first - derivative test.

Step5: Find inflection points

Set $f''(x)=0$, so $12x^{2}-48 = 0$.
$12x^{2}=48$, then $x^{2}=4$, and $x=\pm2$.

Answer:

$f'(x)=4x^{3}-48x + 64$
$f''(x)=12x^{2}-48$
Relative extrema: $x=-4$ (relative minimum), inconclusive at $x = 2$
Inflection points: $x=-2,x = 2$