Calculus function question..!?

Question

Consider the function f(x) = 4 - 2 x^2, -5 (less than or equal to) x (less than or equal to 1. 

The absolute maximum value is: 
and this occurs at x = 

The absolute minimum value is: 
and this occurs at x =

Answer 1

The function f(x) = 4 - 2 x²     -5 ≤ x ≤ 1 .

To find out the absolute maximum / minimum equate the f '(x) = 0 .

 f '(x) = 0 - 2(2)x          [ derivative of  xⁿ = n x^n-1 ]        

 f '(x) =  - 4x 

-4x = 0

x = 0

So the function f(x) = 4 - 2 x² has maximum / minimum at x = 0.

To check whether maximum / minimum find f ''(x)

f '(x) =  - 4x

f ''(x) = - 4 

f ''(x) < 0 , so the function has maximum at x = 0 .

The function has one and only one maximum point so it is the absolute maximum .

To find the maximum value put x = 0 in function .

f(0) = 4 - 2 (0)²

     = 4

So the  absolute maximum value is: 4

Absolute maximum value at x = 0 .

It has no minimum value .