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

Nov 21, 2015

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:

$f \left(x\right) = \left({x}^{2} - 9\right) \left(3 x - 1\right)$, so $f \left(2\right) = 25$.

$f ' \left(x\right) = 2 x \left(3 x - 1\right) + 3 \left({x}^{2} - 9\right)$

$= 9 {x}^{2} - 2 x - 27$

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

Therefore, $f \left(x\right)$ if increasing in some open interval containing $2$.

Nov 21, 2015

The function is growing.

#### Explanation:

And now, using logs, to round out the set

$y = \left(x + 3\right) \left(x - 3\right) \left(3 x - 1\right)$

Take the log of both sides

$\ln \left(y\right) = \ln \left(\left(x + 3\right) \left(x - 3\right) \left(3 x - 1\right)\right)$

Use log properties

$\ln \left(y\right) = \ln \left(x + 3\right) + \ln \left(x - 3\right) + \ln \left(3 x - 1\right)$

Differentiate both sides

${y}^{'} / y = \frac{1}{x + 3} + \frac{1}{x - 3} + \frac{3}{3 x - 1}$

Multiply both sides by $y$

${y}^{'} = \left(x - 3\right) \left(3 x - 1\right) + \left(x + 3\right) \left(3 x - 1\right) + 3 \left(x + 3\right) \left(x - 3\right)$

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

${y}^{'} = \left(2 - 3\right) \left(6 - 1\right) + \left(2 + 5\right) \left(6 - 1\right) + 3 \left(2 + 3\right) \left(2 - 3\right)$
${y}^{'} = - 5 + 7 \cdot 5 - 3 \cdot 5 = - 5 - 15 + 35 = 15$

The function is growing.