# What is the extreme value of g(x)= 3x^2 +5x -12

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g(x) = 3x^2 + 5x -12

We need to find the extreme values.

First we notice that the coefficient of x^2 is positive, then the function has a minimum value.

Now we will find the derivatives zero.

==> g'(x)= 6x + 5 = 0

==> x = -5/6

==> g(-5/6) = 3(25/36) - 25/6 -12

==> g(-5/6)= ( 75 - 150 -432)/ 36 = -507/36 = -169/ 12

**Then the function has a minimum value at f(-5/6) = -169/12**

The extreme values of a function g(x) can be found by solving g'(x) = 0

Here g(x) = 3x^2 +5x -12

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

6x + 5 = 0

=> x = -5/6

For x = -5/6, g(x) = 3*(-5/6)^2 + 5*(-5/6) - 12

=> -169/12

Also g''(x) = 6 which is positive.

**So we have a minimum point at (-5/6, -169/12)**