Question

Is `f(x)=(x+3)(x-3)(3x-1)` increasing or decreasing at `x=2`?

Answer 1

Like all functions it is constant at a value in its domain. Its derivative is continuous and positive at `2`, so it is increasing in some open interval containing `2`.

Explanation:

`f(x)=(x^2-9)(3x-1)`, so `f(2) = 25`.

`f'(x) = 2x(3x-1)+3(x^2-9)`

`= 9x^2-2x-27`

`f'(2) = 5`, and `f'(x)` is continuous, so `f'(x)` is positive for all `x` in some open interval containing `2`.

Therefore, `f(x)` if increasing in some open interval containing `2`.

Answer 2

The function is growing.

Explanation:

And now, using logs, to round out the set

`y = (x+3)(x-3)(3x-1)`

Take the log of both sides

`ln(y) = ln((x+3)(x-3)(3x-1))`

Use log properties

`ln(y) = ln(x+3)+ln(x-3)+ln(3x-1)`

Differentiate both sides

`y^'/y = 1/(x+3) + 1/(x-3) + 3/(3x-1)`

Multiply both sides by `y`

`y^' = (x-3)(3x-1) + (x+3)(3x-1) + 3(x+3)(x-3)`

Put in, `x = 2` and see the value

`y^' = (2-3)(6-1) + (2+5)(6-1) + 3(2+3)(2-3)`
`y^' = -5 + 7*5 - 3*5 = -5 - 15 + 35 = 15`

The function is growing.