SOLUTION: I have this function f(x) = x^4 + 2x^2 - 3 and I am asked to find the max and min values of f on [-1,2] As I understand, to find the max and min I would first take the function

Question 997840: I have this function f(x) = x^4 + 2x^2 - 3 and I am asked to find the max and min values of f on [-1,2]
As I understand, to find the max and min I would first take the functions derivative then set it to zero. However, this problem seems a bit different I know the answers are -3 as the min and 21 as the max but how those were derived I don't know.
Also could someone explain to me how only one of the excluded elements of the interval is allowed to be used as an input for the function? I think 2. Why am I allowed to use 2 as the input but not -1?
My attempt:
f(x) = x^4 + 2x^2 - 3
f'(x) = 4x^3 + 4x
f'(x) = 4x^3 = -4x
f'(x) = 4x^3/(4x) = -4x/(4x)
f'(x) = x^2 = -1
f'(x) = x = -/+sqrt(-1)
this seems very wrong as a possible critical value.
Please explain
Thank you
Answer by ikleyn(52787) (Show Source): You can put this solution on YOUR website!
.
f(x) = .

Take the derivative:

f'(x) = .

Find where the derivative is zero: = .

The only real root is x = 0.

Calculate f(0). It is -3.

The values at the ends x=-1 and x=2 are greater.

So, the minimum is at x=0. The maximum is at x=2.

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