Find the maximum or minimum value of f(x) = 2x^2 + 3x - 5

The extreme point of a function is reached if the value of
x is cancelling out the 1st derivative.

We'll differentiate
the function:

f'(x) = 4x +
3

We'll put f'(x) = 0:

4x + 3
= 0

4x = -3

x =
-3/4

The critical point of the function is x =
-3/4

We'll calculate the 2nd derivative to decide if the
point is maximum or minimum:

f"(x) =
4>0

Since f"(x)>0, then the extreme point of
the function is a minimum point.

We'll substitute x by the
value of the critical point:f(-3/4) = 2*9/16 - 9/4  -
5

f(-3/4) = 18/16 - 36/16 -
80/16

f(-3/4) = -98/16

f(-3/4)
= -49/8

The coordinate of the minimum point
of the function are: (-3/4 ; -49/8).