# Find the minimum values of g(x) =5x^2 - 5x + 7

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Given the parabola g(x) = 5x^2 - 5x + 7.

We need to find the minimum value.

First, we need to find the first derivative.

==> g'(x) = 10x - 5

Now we will calculate the critical values which is the derivative zeros:

==> 10x - 5 = 0

==> 10x = 5

==> x = 1/2.

Since the sign of x^2 is positive, then we know that the function has a minimum value at x= 1/2

Then the function has a minimum value at x= 1/2

==> g(1/2) = 5(1/2)^2 - 5(1/2) + 7

= 5/4 - 5/2 + 7

= ( 5 - 10 + 28)/4

= 23/4

**Then, the minimum values is: f(1/2) = 23/4**

We can find the point at which minimum and maximum values of any function occurs equating the derivative of the function to 0, and finding the corresponding value of the function.

Thus:

g'(x) = 10x - 5

Equating this to 0:

10x - 5 = 0

==> 10x = 5

==>x = 1/2

Substituting this value of x in the function g(x):

g(1/2) = 5(1/2)^2 - 5*1/2 + 7

= 5/4 - 5/2 + 7 = 23/4 = 5.75

We can check if 5.75 is the maximum or minimum value of g(x) comparing it with any other value of x, say 1.

g(1) = 5*1^2 - 5*1 + 7 = 7

This is more than g(1/2) = 5.75

Therefore, minimum value of g(x) is 5.75.

To find the minimum value of g(x) = 5x^2-5x+7.

g(x) = 5(x^2-x)+7

g(x) = 5{x^2-2*(1/2)x +(1/2)^2} -5(1/2)^2+7. We added 5(1/2)^2 and subtracted 5(1/2)^2 .

g(x) = 5(x-1/2)^2-5/4+7.

g(x) = 5(x-1/2)^2 + (28-5)/4

g(x) = 5(x-1/2)^2 + 23/4.

Therefore g(x) > = 23/4 as, 5(x-1/2)^2 is always > 0 for all x and for x= 1/2, g(1/2) = 23/4 is minimum as 5(x-1/2)^2 = 0 for x = 1/2.