Which is the minimal value for the function f(x)=x^2+2x?
To find the minimum value for the function, first we need to determine the critical or extreme values for the function.
f' = 2x+2
==> x =-1 is an extreme value for f
To determine if x=-1 is a minimum or maximum value we need to calculate f'' . If f'' is positive, then x=-1 is a minimum.
f'' = 2 which is positive
Then f(x) has minimum value at x=-1
The minimum and maximum values of any function f(x) occurs at point where:
f'(x) = 0
f(x) = x^2 + 2x
f'(x) = 2x + 2
Equating f'(x) to 0 we get:
2x + 2 = 0
2x = -2
x = -2/2 = -1
For x = -1, value of f(x) is given by:
f(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1
We find that any value of x other than -1 gives value of f(x) greater than -1. Therefore minimum value of the given function is -1.
f(x) = x^2+2x. To find the minimal value .
f(x) = x^2+2x.
Then the extreme values are determined by equating f'(x) = 0 and solving for x.
f'(x) = (x^2+2x)' = 2x+2. Equating x+2 to zero, we get: x+2 = 0. Or
Now if f"(-2 ) = 2 which is positive. Therfore x^2+2x is minimum for x=-2. Or
f(-2) = (-2)^2+2*2 = -4+4 = 0 is the minimum value.
To calculate the extreme value for a function, minimum or maximum, we have to differentiate the function,which, in this case, is:
f'(x)= (x^2)' + (2x)'
f'(x)= 2*x + 2*1
Now, we'll calculate the root of the first derivative, this value being the value for the function f has the minimum value.
We'll divide by 2:
The minimum value of the function is
f(-1) = (-1)^2 + 2*(-1)
f(-1) = 1 - 2
f(-1) = -1