Let f be the function given by f(x) = x+4(x−1)(x+3) on the closed interval [−5,5]. on which closed interval is the function f guaranteed by the extreme value theorem to have an absolute maximum and an absolute minimum?

For given function, absolute maxima at x = 5 and absolute minima at

Absolute maximum value is 133 and absolute minimum value is

Given function is,

To find critical points, Differentiate above function and equate with zero.

We have critical points are,

From above, it is clear that is absolute maximum value and is absolute minimum value.

Thus, For given function, absolute maxima at x = 5 and absolute minima at

Absolute maximum value is 133 and absolute minimum value is

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Step-by-step explanation:

f(x)=x+4(x^2+2x-3)=4x^2+9x-12

f'(x)=8x+9

f'(x)=0,gives x=-9/8

f(-5)=-5+4(-5-1)(-5+3)=-5+4*-6*-2=43

f(-9/8)=-9/8+4(-9/8-1)(-9/8+3)

=-9/8+4*-17/8*15/8

=-9/8-255/16

=-273/16=-17 1/16

f(5)=4*5^2+9*5-12=100+45-12=133

absolute maximum=133

absolute minimum=-17 1/16

well what you have to do is divide 84 by 6 so theanswer

step-by-step explanation:

84 divided by 6 = 14

and that lightbulb changing party must be freaking lit!

10.4 inches

step-by-step explanation:

2x2x2.6