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

Apr 4, 2016

It is impossible to determine because the gradient is a complex number.

#### Explanation:

The question is essentially asking whether $f ' \left(x\right) < \text{or} > 0$ at x = -1.

To answer this we need to find the derivative of $f \left(x\right)$.

Start by taking out the exponent inside the logarithm

$- x \ln \left(2 {x}^{2}\right) = - 2 x \ln \left(2 x\right)$

You can differentiate this using the product rule, where

$f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$
$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + h ' \left(x\right) g \left(x\right)$

When $g \left(x\right) = - 2 x$ and $h \left(x\right) = \ln \left(2 x\right)$,

$g ' \left(x\right) = - 2$

$h \left(x\right) = \ln 2 + \ln x$
$h ' \left(x\right) = \frac{d}{\mathrm{dx}} \ln 2 + \frac{d}{\mathrm{dx}} \ln x$

And, since $\ln 2$ is a constant,

$h ' \left(x\right) = \frac{1}{x}$

Inserting both of these values into the product rule equation,

$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + h ' \left(x\right) g \left(x\right)$
$f ' \left(x\right) = - 2 \cdot \ln 2 x + \frac{- 2 x}{x}$

Substituting $x = - 1$,

$f ' \left(- 1\right) = - 2 \left(0.7 + \pi i\right) + \frac{2}{-} 1$
$f ' \left(- 1\right) = - 3.4 - 2 \pi i$

It is therefore impossible to determine whether it is an increasing or decreasing function at $x = - 1$ because imaginary numbers ($i = \sqrt{-} 1$) cannot easily be dealt with arithmetically.

Apr 5, 2016

Decreasing.

#### Explanation:

To determine if a function is increasing or decreasing at a certain point, we look at the sign of its derivative at that point.

If the function's derivative is $> 0$ at a point, then it is increasing. Similar logic applies in that a negative $\left(< 0\right)$ value of the derivative is indicative of the function decreasing at such a point.

So, we must find the derivative of $f \left(x\right)$. To do so, we will need to use the product rule. Applying the product rule gives the derivative to be:

$f ' \left(x\right) = \ln \left(2 {x}^{2}\right) \frac{d}{\mathrm{dx}} \left(- x\right) + \left(- x\right) \frac{d}{\mathrm{dx}} \left(\ln \left(2 {x}^{2}\right)\right)$

We can find the internal derivatives so that we can simplify:

$\frac{d}{\mathrm{dx}} \left(- x\right) = - 1$

To find the next derivative, we must use the chain rule. Applied specifically to the natural logarithm function, we see that

$\frac{d}{\mathrm{dx}} \left(\ln \left(u\right)\right) = \frac{1}{u} \cdot u '$

Thus, we see that

$\frac{d}{\mathrm{dx}} \left(\ln \left(2 {x}^{2}\right)\right) = \frac{1}{2 {x}^{2}} \cdot \frac{d}{\mathrm{dx}} \left(2 {x}^{2}\right) = \frac{1}{2 {x}^{2}} \cdot 4 x = \frac{2}{x}$

Plugging these back in, we see that

$f ' \left(x\right) = \ln \left(2 {x}^{2}\right) \cdot \left(- 1\right) - x \left(\frac{2}{x}\right)$

$= - \ln \left(2 {x}^{2}\right) - 2$

So, to determine if the function is increasing or decreasing at $x = - 1$, we must find $f ' \left(- 1\right)$.

f'(-1)=-(ln(2(-1)^2)-2=-ln(2)-2

We don't need a calculator to determine that this is negative, hence the function is decreasing at $x = - 1$.

(For the sake of completeness, you may want to note that -ln(2)-2approx-2.693<0).

We can verify this claim by checking a graph of $f \left(x\right) = - x \ln \left(2 {x}^{2}\right)$:

graph{-xln(2x^2) [-5.45, 5.647, -2.014, 3.534]}

At $x = - 1$, the function is indeed decreasing.