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

##### 2 Answers

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

#### Explanation:

The question is essentially asking whether

To answer this we need to find the derivative of

Start by taking out the exponent inside the logarithm

You can differentiate this using the product rule, where

When

And, since

Inserting both of these values into the product rule equation,

Substituting

It is therefore impossible to determine whether it is an increasing or decreasing function at

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

So, we must find the derivative of

#f'(x)=ln(2x^2)d/dx(-x)+(-x)d/dx(ln(2x^2))#

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

#d/dx(-x)=-1#

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

#d/dx(ln(u))=1/u*u'#

Thus, we see that

#d/dx(ln(2x^2))=1/(2x^2)*d/dx(2x^2)=1/(2x^2)*4x=2/x#

Plugging these back in, we see that

#f'(x)=ln(2x^2)*(-1)-x(2/x)#

#=-ln(2x^2)-2#

So, to determine if the function is increasing or decreasing at

#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

We can verify this claim by checking a graph of

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

At