We have to find the maximum value of f(x)= -2x^2 -6x + 3.

To get the result we find the derivative of f(x) and equate f'(x) to zero. We then solve for x and use that to find the value of f(x).

f(x)= -2x^2 -6x + 3

=> f'(x) =...

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We have to find the maximum value of f(x)= -2x^2 -6x + 3.

To get the result we find the derivative of f(x) and equate f'(x) to zero. We then solve for x and use that to find the value of f(x).

f(x)= -2x^2 -6x + 3

=> f'(x) = -4x - 6

-4x - 6 = 0

=> x = -6/4 = -3/2

Also, we see that f''(x) = -4 which is negative ensuring that the solution we get is the maximum value.

f(-3/2) = -2*(-3/2)^2 - 6(-3/2) + 3

=> -2*9/4 + 9 + 3

=> 15/2

**The required maximum value is 15/2**

We have the function:

f(x) = -2x^2 - 6x + 3

We need to find the maximum value of f(x).

First we look at the coefficient of x^2 which is negative.

Then the fucntion has a maximum point.

Now we will find the first derivative and determine the zeros.

==> f'(x) = -4x - 6 = 0

==> -4x = 6

==> x = -6/4 = -3/2= -1.5

Now we will find f(-3/2)

==> f(-3/2) = -2(-3/2)^2 - 6(-3/2) +3 = -18/4 + 18/2 + 3 = (-18+36 + 12)/4 = 30/4 = 15/2 = 7.5

**Then,**** the maximum value of the function f(x) is the point **

**( -1.5, 7.5)**