# Which is the minimal value for the function f(x)=x^2+2x?

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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

2x+2= 0

==> 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

Given:

f(x) = x^2 + 2x

Therefore:

f'(x) = 2x + 2

Equating f'(x) to 0 we get:

2x + 2 = 0

2x = -2

Therefore:

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 .

Solution:

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

x =-2.

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

f'(x)= 2x+2

Now, we'll calculate the root of the first derivative, this value being the value for the function f has the minimum value.

2x+2=0

2x=-2

We'll divide by 2:

x=-1

The minimum value of the function is

f(-1) = (-1)^2 + 2*(-1)

f(-1) = 1 - 2

**f(-1) = -1**