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

To find the minimum value , first we identify the sign of x^2, since the sign if positive, that means the function has a minimum value.

First we will calculte the critical values of the function which are the first derivative's zeros.

==> f'(x) = 6x + 5

==> 6x + 5= 0

==> 6x = -5

==> x= -5/6

Now we know that the function has a minimum values at x= -5/6

To find the value we will substitute in f(x);

==> f(-5/6) = 3*(-5/6)^2 + 5(-5/6) - 3

= 3*25/36 - 25/6 - 3

= (75 - 150 - 108)/36

= -183/36 = -5.08

**Then the function has a minimum value at f(-5/6) = -5.08**

To find the minimum value of f(x) = 3x^2+5x-3.

We know by calculus that f(x) is minimum if there is a solution x = c , for f'(c) = 0 and f"(c) < 0.

So we differentiate f'(x) and set it to zero and find the solution say x = c, and f''(c) < 0.

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

f'(x) = 3*2x+5 = 6x+5

f'(x) = 0 gives, 6x+5 = 0

6x= -5

x = -5/6.

f"(x) = (6x+5)' = 6 > 0 for all x.

Therefore for x = c = -5/6 f(-5/6) is minimum.

So f(-5/6) = 3(-5/6)^2+5(-5/3)-3 = 3*25/36 -25/3 -3 = 25/12 -25/3 -3 = - 9.25 is the minimum value at x= -5/6.