Find the maximum value of f(x) = -2x^2+12x-14?

Answer 1

f(-b/2a)=f(3)=-2*9+36-14=-18+36-14=4=maximum

Answer 2

Take the derivative of the function and set it equal to zero. So,

f'(x) = -4x + 12 = 0

Solving for this, you get that x = 3 at the maximum, so plug this back into your function. You get:

f(3) = -2(3)^2 + 12(3) -14 = -18+36-14 = 4.

Therefore, the maximum is 4 at x=3.

Answer 3

- 2x^2 + 12x - 14
= -2( x^2 - 6x + 7 )
= -2( (x - 3)^2 - 2 )
= 2(2 - (x - 3)^2).

This has a maximum value of 4 when (x - 3)^2 = 0.

Answer 4

4

Answer 5

maximum (and minimum) values can be found using the first derivitive set equal to 0.

taking the first derivitive, we get
f '(x) = -4x + 12
set that = 0 and solve for x and we get
-4x = -12
x = 3

now we take that three and plug it back into the function to get the maximum value

-2(3)^2 + 12(3) - 14 =
-18 + 36 - 14 = 36 - 32 = 4

the maximum value of the above function is 4.

Answer 6

f(x) = -2x^2 +12x - 14

ok, so you will take the first derivative of this function which is:

f'(x) = -4x + 12

then you solve for x:

-4x + 12 = 0

-4x = -12

x = 3

then you will plug this back into the original function:

f(3) = -2(3)^2 + 12(3) -14
= -18 + 36 - 14
= 4

thus the max. value of the function is 4.

Answer 7

dy/dx = -4x+12
Set = 0 getting x = 3
f(3) = -2(3^2) +12*3 -14
f(3) = -18 +36 -14 = 4
max value = 4

Answer 8

ehm.... do derivative of that, set it to 0 and calculate the values for x

Answer 9

about three... I think. I didn't do the math, I just looked at it.