Is f(x)=ln(2x^2-1) increasing or decreasing at x=-1?

Apr 9, 2016

$f \left(x\right)$ is decreasing.

Explanation:

To determine if a function is increasing or decreasing, take the derivative and evaluate it at the $x$-value in question. If the derivative is positive, that means the slope is positive (because derivative is the slope), and therefore the function is increasing. If the derivative is negative, then the slope is negative and therefore the function is decreasing.

We start at by finding $f ' \left(x\right)$. We know the derivative of $\ln x$ is $\frac{1}{x}$; to find the derivative of $\ln \left(2 {x}^{2} - 1\right)$, we need to apply that plus the chain rule:
$f \left(x\right) = \ln \left(2 {x}^{2} - 1\right)$
$f ' \left(x\right) = \frac{1}{2 {x}^{2} - 1} \cdot 4 x \to$ $\frac{d}{\mathrm{dx}} \left(2 {x}^{2}\right) = 4 x$, so we multiply it by the whole derivative (because the chain rule requires us to do so)
$f ' \left(x\right) = \frac{4 x}{2 {x}^{2} - 1}$

We finish off by evaluating $f ' \left(- 1\right)$ to see if $f \left(x\right)$ is increasing or decreasing at $x = - 1$:
$f ' \left(- 1\right) = \frac{4 \left(- 1\right)}{2 {\left(- 1\right)}^{2} - 1}$
$f ' \left(- 1\right) = - \frac{4}{1} = - 4$

Because the slope at $x = - 1$ is negative, the function is decreasing. You can confirm this by looking at the graph of $\ln \left(2 {x}^{2} - 1\right)$ - you'll see it is indeed decreasing. graph{ln(2x^2-1) [-10, 10, -5, 5]}