Help with calculus!!?

Question

i need help!
how do i do this??

Consider the function f(x) = 2 x^3 + 12 x^2 - 126 x + 8, -7≤ x≤4. This function has an absolute minimum value equal to___?

Answer 1

The absolute minimum value is equal to -208

The solution is shown as follows:

First, find the first derivative of the function f(x):

f'(x) = 6x^2 + 24x - 126



After finding the first derivative, try if you can simplify it. As you can see, it can be simplified to...

x^2 + 4x - 21

Which can be factored out as (x+7)(x-3).



Letting the factors equal to ZERO, (the rule in the first derivative test), or for the simple reason that a curve can be at its MAXIMUM or its MINIMUM when it's slope (first derivative) is equal to ZERO. That's the reason why we let the factors be equal to zero. To either get the MAXIMUM or the MINIMUM. Remember that part, okay? =) Anyway, going on, we get...

(x+7)(x-3) = 0

Hence, we get two values for x.

Value #1: x = -7
Value #2: x = 3

Are these two values inside our GIVEN INTERVAL?

YES! They are part of -7≤ x ≤ 4. Hence, both of them are valid values.



Substituting these two values in the ORIGINAL EQUATION, one of them will give us the MAXIMUM, and the other is the MINIMUM. But which is it?

If we try x=-7, we get:

f(-7) = 2(-7)^3 + 12(-7)^2 - 126(-7) + 8 = 1137

It's too big a value for it to be the absolute MINIMUM value, let's try x=3:

f(3) = 2(3)^3 + 12(3)^2 - 126(3) + 8 = -208

Yes! -208 is the ABSOLUTE MINIMUM VALUE.



An explanation:
If you try substituting all of the numbers in the given interval -7≤ x ≤ 4, only x=3 will give the SMALLEST value. That's why it's the ABSOLUTE MINIMUM VALUE.

God bless on your homework or exams! =)

Answer 2

f(x) = 2 x^3 + 12 x^2 - 126 x + 8
f(-7) = 2 (-7)^3 + 12 (-7)^2 - 126 (-7) + 8
f(-7) = 792
f(4) = 2 (4)^3 + 12 (4)^2 - 126 (4) + 8
f(4) = -176
f'(x) = 6 x^2 + 24 x - 126 = 0 for criticals:
x^2 + 4x - 21 = 0
(x + 7)(x - 3) = 0
x = -7, 3
f(3) = 2 (3)^3 + 12 (3)^2 - 126 (3) + 8
f(3) = - 208

Answer 3

the roots of the derivative are 3 and -7 and x can go from -7 to 4. we substitute the values on the second derivative to know if they are max or min. and we get the maximum is at -7 and the minimum is at 3.