# `f(x) = 6x - x^2` Find all relative extrema, use the second derivative test where applicable.

### Textbook Question

Chapter 3, 3.4 - Problem 31 - Calculus of a Single Variable (10th Edition, Ron Larson).
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thilina-g | College Teacher | (Level 1) Educator

Posted on

`f = 6x -x^2`

`f' = 6-2*x`

`f' = 2(3-x)`

At extrema, f'=0, this give `3- x = 0 therefore x =3 `

At x = 3, there is a local extreme point. At x = 3,

`f = 6*3-3^2 = 18-9 = 9`

Second derivative test,

`f'' = 2(0-1) = -2`

`f'' lt 0` for all `x` , therefore at `x =3, f'' lt 0.`

This suggests, the extrem point at x = 3 is a maximum.

Therefore, there is a local maximum point of (3,9)

Educator Approved

loves2learn | (Level 3) Salutatorian

Posted on

Find the first derivative.

Given,

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

`f'(x)=6-2x `

Now we need to find the value(s) of x that make the first derivative zero to find the critical points.

`0=6-2x `

`x=3 ` (critical point)

This will be a relative extrema if it changes sign, so find 2 values around it to test using the first derivative test, like 2 and 4

`f'(4)=-2 `

`f'(2)=2 `

Since the sign changes then the point `(3,9) ` is a relative extrema