Question

find the absolute maximum and minimum values of each function over the indicated interval, and indicate the x - values at which they occur
$f(x)=15 x^{4}-4 x^{3},-2,2$
the absolute maximum value is \\( \square \\) at \\( x= \\)
(use a comma to separate answers as needed )
the absolute minimum value is \\( \square \\) at \\( x= \\)
(use a comma to separate answers as needed )

Explanation:

Step1: Find the derivative of the function

The function is \( f(x)=15x^{4}-4x^{3}\). Using the power rule \((x^{n})^\prime = nx^{n - 1}\), the derivative \(f^\prime(x)=(15x^{4}-4x^{3})^\prime=15\times4x^{3}-4\times3x^{2}=60x^{3}-12x^{2}=12x^{2}(5x - 1)\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(12x^{2}(5x - 1)=0\).
By the zero - product property, \(x = 0\) or \(x=\frac{1}{5}\). Both \(x = 0\) and \(x=\frac{1}{5}\) are in the interval \([-2,2]\).

Step3: Evaluate the function at the critical points and endpoints

• When \(x=-2\): \(f(-2)=15\times(-2)^{4}-4\times(-2)^{3}=15\times16+4\times8=240 + 32=272\).
• When \(x = 0\): \(f(0)=15\times0^{4}-4\times0^{3}=0\).
• When \(x=\frac{1}{5}\): \(f(\frac{1}{5})=15\times(\frac{1}{5})^{4}-4\times(\frac{1}{5})^{3}=15\times\frac{1}{625}-4\times\frac{1}{125}=\frac{15 - 20}{625}=-\frac{1}{125}=- 0.008\).
• When \(x = 2\): \(f(2)=15\times2^{4}-4\times2^{3}=15\times16-4\times8=240-32 = 208\).

Answer:

The absolute maximum value is \(272\) at \(x=-2\).
The absolute minimum value is \(-\frac{1}{125}\) at \(x=\frac{1}{5}\).