f= 5x^2 - 15 x + 1

First we need to determine the first derivative f' and find critical values:

f'(x) = 10x - 15

Now to calculate critical value we need to find f'(x)'s zeros:

=> 10x - 15 = 0

==> 10x = 15

==: x = 15/10 = 3/2

Now f(3/2) = 5(3/2)^2 -15(3/2) +1

= 45/4 - 45/2 + 1

= (45-90 +4) / 4 = -41/4

Then the function has a minimmum value at the point (3/2, -41/4)

f(x) = 5x^2-15x+1

To find the minimum of f(x).

Solution:

f(x) = 5x^2-15x+1

f(x)= 5(x^2-3x) +1

f(x) = 5[x^2-3x +(3/2)^2] + 1 - 5(3/2)^2 .

f(x) = 5(x-3/2)^2 + {4-45]/4

f(x) = 5(x-3/2)^2 - 41/4 > -41/4 when x = 3/2 as 5(x-3/2)^2 is always a positive quantity.

Therefore f(3/2) = -41/4 is the minimum value when x= 3/2.

We'll choose to do the derivative test to calculate the extreme value of a function:

f'(x) = (5x^2 - 15*x + 1)'

f'(x) = 10x - 15

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.

10x - 15 = 0

We'll add 15 both sides:

10x = 15

We'll divide by 5 both sides:

2x = 3

We'll divide by 2 both sides:

x = 3/2

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

f(3/2) = 5*(3/2)^2 - 15*(3/2) + 1

f(3/2) = 45/4 - 45/2 + 1

We'll eliminate like terms:

f(3/2) = -45/4 + 1

**f(3/2) = -41/4**

**The minimum value of the function is -41/4.**