Find the relative maximum of -x2 plus 6x-4?

Answer 1

f(x) = -x2 + 6x - 4

∴ f'(x) = -2x + 6

Solve for f'(x) = 0:

0 = -2x + 6

x = 3

So the peak occurs at the point where x = 3. Now you can calculate the value of f(3)

f(3) = -32 + 6(3) - 4

∴ f(3) = -9 + 18 - 4

∴ f(3) = 5

So the maximum value of that function is 5, and it occurs when x = 3.

To prove that it's a maximum, take the second derivative:

f'(x) = -2x + 6

∴ f''(x) = -2

Because it's negative, we know that this is the peak of the curve.