What is the extreme value of  y= 3x^2 -3x +5, Is is max. or min?

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To find the extreme value of y we need to find the first derivative and equate it to zero to solve for x.

y = 3x^2 - 3x + 15

y' = 6x - 3

equating it to 0

6x - 3 = 0

=> x = 3/6

=> x = 1/2

For x = 1/2, y = 3/4 - 3/2 + 5 = 17/4

The second derivative of y is 6. This is positive for x = 1/2.

Therefore we have a minimum at x = 1/2.

The required extreme value is 17/4, and it is a minimum.

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Given the equation:

y= 3x^2 - 3x + 5

We need to find the extreme value.

First we need to determine the first derivative.

==> y' = 6x -3

Now we will calculate the critical value.

==> 6x -3 = 0  ==> x = 1/2

Now we will determine the values of y(1/2)

==> y(1/2) = 3(1/4) - 3(1/2) + 5 = 3/4 - 3/2 + 5 = ( 3 - 6 + 20)/4 = 17/4

==> y(1/2) = 17/4

Since the coefficient of x^2 is positive, then the parabola is facing up.

Then the parabola has a minimum point at x= 1/2.

Then the extreme values for y is the point ( 1/2, 17/4) and it is a minimum.

Approved by eNotes Editorial Team

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