f(x) = x^2 - mx + 3

Let us calculate the first derivative and find critical values:

f'(x) = 2x -m

2x -m = 0

==> x = m/2 (criticl value)

Then the exrtreme value is f(m/2) = -1

==> f(m/2) = (m/2)^2 - m(m/2) + 3

-1= m^2/4 - m^2/2 + 3

Multiply by 4 :

==> -4 = m^2 - 2m^2 + 12

==> -4 = -m^2 + 12

==> m^2 = 16

==> m = +-4

To calculate the extreme value of a function, we have to calculate it's derivative.

f'(x) = (x^2 - m*x + 3)'

f'(x) = 2x - m

When the first derivative is cancelling, then the function has an extreme value, in this case, a minimum.

2x - m = 0

2x = m

x = m/2

But, from enunciation, f(m/2) = -1.

f(m/2) = m^2/4 - m^2/2 + 3

-1 = (m^2 - 2m^2 + 12)/4

-4 = -m^2 + 12

We'll move -m^2 to the left side:

m^2 - 4 = 12

We'll add 4 both sides:

m^2 = 16

m1 = -sqrt 16

**m1 = -4**

**m2 = +4**

To find the x for which minimum for f(x) = x^2-mx+3 is -1.

Solution:

f(x) = x^2-2mx/2+(m/2)^2 -(m/2)^2+3 , as m/2)^2 is added and subtracted.

= (x-m/2)^2 -m^2/4 +3 is minimum and equalt to -m^2/4 +3 when

x =-m/2 .

So -m^2/4+3 = -1 at minimum

-m^2/4 = -1-3 = -4

m^2 = 4*4 =16Or

m = sqrt16.

Or

m = 4 or m = -4