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)=4x^3 - 48x + 64
f(x)=12x^2 - 48
b. complete
the function f(x) has a relative minimum at x =

Explanation:

Step1: Find critical points

Set $f^{\prime}(x)=4x^{3}-48x + 64 = 0$. Factor out 4: $4(x^{3}-12x + 16)=0$, so $x^{3}-12x + 16=0$. By trial - and - error, we find that $x = 2$ is a root. Then we can factor $x^{3}-12x + 16=(x - 2)(x^{2}+2x - 8)=(x - 2)(x - 2)(x+4)=(x - 2)^{2}(x + 4)=0$. The critical points are $x=-4$ and $x = 2$.

Step2: Use the second - derivative test

We have $f^{\prime\prime}(x)=12x^{2}-48$.
For $x=-4$, $f^{\prime\prime}(-4)=12\times(-4)^{2}-48=12\times16 - 48=192-48 = 144>0$. Since $f^{\prime}(-4)=0$ and $f^{\prime\prime}(-4)>0$, the function has a relative minimum at $x=-4$.
For $x = 2$, $f^{\prime\prime}(2)=12\times2^{2}-48=48 - 48=0$. The second - derivative test is inconclusive for $x = 2$.

Answer:

-4