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

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)

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