Question

How do you find the maximum or minimum of `f(x)=3-x^2-6x`?

Answer 1

`(-3,12)`

Explanation:

Since this is a parabolic equation and the value of `a<0` it open downwards so it have a absolute maxima. The maximum point is determined by `x_(max) = -b/(2a)` where b and a are coefficients. This formula is developed using differential calculus.
So,
`x_(max) = -b/(2a)`
`=-(-6)/(2×(-1))`
`=6/(-2)`
`:. X_(max) = -3`.
Now evaluate `f (-3)` get the maximum point.

Answer 2

It's Maximum

Explanation:

The second derivative demonstrates whether a point with zero first derivative is a maximum, minimum or an inflexion point.

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

`f'(x) = (d/(dx)) (3 - 6x - x^2 = -6 - 2x`

`-6 - 2x = 0 " or" x = -3`

f"(x) `= (d/(dx)) (-2x - 6 ) = -2`

Since f"(x) is negative, it is maximum.

graph{(-x^2 - 6x + 3) [-11.21, 8.79, -13.32, -3.32]}