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

### 3 Answers | Add Yours

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

At the point of maximum or minimum f'(x) = 0

f'(x) = 4x + 3

4x + 3 = 0

=> x = -3/4

f''(x) = 4, which is positive.

Therefore we get a minimum value at x = -3/4

f(-3/4) = 2*(-3/4)^2 - 3*(3/4) - 5

=> -49/8

**The minimum value of the function is -49/8**

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

First we notice that the function has a positive coefficient for x^2.

Then the fucntion has a minimum values.

Now we will find the derivatives zero.

==> f'(x) = 4x +3 = 0

==> 4x = -3

==> x = -3/4

==> f(-3/4)= 2(-3/4)^2 + 3(-3/4) -5

= 2* 9/16 - 9/4 - 5

= (18- 36 - 80)/16 = -98/16 = -49/8.

**==> The minimum values is f(-3/4) = -49/8.**

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).**