find the absolute maximum and minimum values of each function over the indicated interval, and indicate the…

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

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

Answer

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}).