(1 point)\nfor the function f(x)=x³ - 3x² - 5 on the interval -1,4, find the absolute extrema, as well as…

(1 point)\nfor the function f(x)=x³ - 3x² - 5 on the interval -1,4, find the absolute extrema, as well as all values of x where they occur.\nabsolut?s maximum of at x = \nabsolute minimum of at x = \n(if an extremum occurs at multiple values of x, enter them as a comma - separated list like 5, 6, 8.)\nnote: you can earn partial credit on this problem.

(1 point)\nfor the function f(x)=x³ - 3x² - 5 on the interval -1,4, find the absolute extrema, as well as all values of x where they occur.\nabsolut?s maximum of at x = \nabsolute minimum of at x = \n(if an extremum occurs at multiple values of x, enter them as a comma - separated list like 5, 6, 8.)\nnote: you can earn partial credit on this problem.

Answer

Explanation:

Step1: Find the derivative

$f'(x)=3x^{2}-6x = 3x(x - 2)$

Step2: Find critical points

Set $f'(x)=0$, then $3x(x - 2)=0$. So $x = 0$ and $x=2$ are critical points.

Step3: Evaluate the function at critical and end - points

Evaluate $f(x)$ at $x=-1,0,2,4$. $f(-1)=(-1)^{3}-3(-1)^{2}-5=-1 - 3-5=-9$ $f(0)=0^{3}-3\times0^{2}-5=-5$ $f(2)=2^{3}-3\times2^{2}-5=8 - 12-5=-9$ $f(4)=4^{3}-3\times4^{2}-5=64 - 48-5=11$

Answer:

Absolute maximum of $11$ at $x = 4$ Absolute minimum of $-9$ at $x=-1,2$