Question

consider the function ( f(x)=3 x^{2}-4 x^{3}+4 x + 8 ) on the interval ( (-infty,infty) ). for each of the following, enter dne if a value does not exist. this function has an absolute minimum value equal to which is attained at ( x = ). this function has an absolute maximum value equal to which is attained at ( x = ).

Explanation:

Step1: Find the first derivative

The function is \(f(x)=3x^{2}-4x^{3}+4x + 8\).
Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we get \(f^\prime(x)=6x-12x^{2}+4\).
Set \(f^\prime(x) = 0\), so \(12x^{2}-6x - 4=0\).
Dividing by \(2\) gives \(6x^{2}-3x - 2=0\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 6\), \(b=-3\), \(c=-2\)), we have \(x=\frac{3\pm\sqrt{(-3)^{2}-4\times6\times(-2)}}{2\times6}=\frac{3\pm\sqrt{9 + 48}}{12}=\frac{3\pm\sqrt{57}}{12}\).

Step2: Find the second derivative

Differentiate \(f^\prime(x)=6x-12x^{2}+4\) to get \(f^{\prime\prime}(x)=6-24x\).
For \(x_1=\frac{3+\sqrt{57}}{12}\), \(f^{\prime\prime}(x_1)=6-24\times\frac{3+\sqrt{57}}{12}=6-(6 + 2\sqrt{57})=- 2\sqrt{57}<0\), so \(x=\frac{3+\sqrt{57}}{12}\) is a local maximum.
For \(x_2=\frac{3-\sqrt{57}}{12}\), \(f^{\prime\prime}(x_2)=6-24\times\frac{3-\sqrt{57}}{12}=6-(6 - 2\sqrt{57})=2\sqrt{57}>0\), so \(x=\frac{3-\sqrt{57}}{12}\) is a local minimum.
Since \(\lim_{x
ightarrow-\infty}f(x)=\lim_{x
ightarrow-\infty}(-4x^{3}+3x^{2}+4x + 8)=\infty\) and \(\lim_{x
ightarrow\infty}f(x)=\lim_{x
ightarrow\infty}(-4x^{3}+3x^{2}+4x + 8)=-\infty\), there is no absolute maximum and no absolute minimum.

Answer:

The function has no absolute minimum value (enter DNE) and no absolute maximum value (enter DNE).