# Verify if the function f=x^2-4x+4 has an extreme point!Verify if the function f=x^2-4x+4 has an extreme point!

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To determine the maxima and minima of the function we can use the first derivative test:

First Derivative Test

If f'(x)>0 for all xEI, then f is increasing on I

If f'(x)<0 for all xEI, then f is decreasing on I

Given the equation: x^2-4x+4

f'(x) = 2x - 4

First we need to find the case at 0:

f'(x) = 2x-4 = 0 --> 2x=4 --> x = 2

Now we need to know whether the function is increasing or decreasing before 2, and after 2.

For x=2 to be a maximum the function needs to be increasing for x<2 and decreasing for x>2.

For it to be a minimum the function needs to be decreasing for x<2 and increasing for x>2.

To test this we pick a value above 2 and below 2:

x= 3 --> f'(x) = 2(3) - 4 = 2 (positive = increasing)

x=1 --> f'(x) = 2(1) - 4 = -2 (negative = decreasing)

Therefore there is a minima in this function at x=2.

f(x)= x^2 - 4x + 4

First let us find f'(x)

f'(x) = 2x - 4

==> To find critical values, we need to calculate f'(x) zeros.

2x - 4 = 0

==> 2x = 4

==> x= 2

Now let us substitute in f(x)

f(2) = 2^2 - 4(2)+ 4 = -8

Then the function has a minimmum value at (2, -8)

The extreme points of x^2-4x+4 is given by:

f'(x) = 0.

f('x) = (x^2-4x+4)'

=2x-4. setting this to zero, we get 2x-4 = 0

2x =4

x =2 is the extreme point.

To verify if the function has an extreme point, we have to calculate it's derivative .

f'(x) = (x^2-4x+4)'

f'(x) = 2x - 4

When the first derivative is cancelling, then the function has an extreme value, in this case, a minimum, because the coefficient of x^2 is positive.

2x - 4 = 0

2x = 4

We'll divide by 2 both sides:

x = 2

Now, we'll calculate the minimum value of the function:

f(2) = 2^2 - 4*2 + 4

f(2) = 0

The function has an extreme value when x=2.

Another method of calculating the first derivative:

f(x) = (x-2)^2

f'(x) = 2(x-2)

2(x-2) = 0

We'll divide by 2 both sides:

x-2 = 0

x = 2