Where does the extreme value of y = -3x^2 + 2x + 5 lie?

Monday, October 1, 2012

Given the function:

y=
-3x^2 + 2x + 5

To find the extreme values, first we will
find the derivative y'.

==> y' = -6x +
2

Now we will find the critical values which is the
derivatives zeros.

==> -6x + 2 =
0

==> -6x = -2 ==> x= 2/6 =
1/3

Then, the function has extreme value when x=
1/3

==> y(1/3) = -3(1/3)^2 + 2(1/3) +5 = -1/3 + 2/3
+ 15/3 = 16/3

We notice that the sign of x^2 is negative.
Then, the function has a maximum values at y(1/3) =
16/3.

Then,the maximum values is the point
(1/3, 16/3)

Notes