Question

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

Answer 1

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

Explanation:

The question is essentially asking whether `f'(x) < "or" > 0` at x = -1.

To answer this we need to find the derivative of `f(x)`.

Start by taking out the exponent inside the logarithm

`-xln(2x^2) = -2xln(2x)`

You can differentiate this using the product rule, where

`f(x) = g(x) * h(x)`
`f'(x) = g'(x)h(x) + h'(x)g(x)`

When `g(x) = -2x` and `h(x) = ln(2x)`,

`g'(x) = -2`

`h(x) = ln2 + lnx`
`h'(x) = d/dx ln2 + d/dx lnx`

And, since `ln2` is a constant,

`h'(x) = 1/x`

Inserting both of these values into the product rule equation,

`f'(x) = g'(x)h(x) + h'(x)g(x)`
`f'(x) = -2 * ln2x + (-2x)/x`

Substituting `x = -1`,

`f'(-1) = -2(0.7 + pii) + 2/-1`
`f'(-1) = -3.4 - 2pii`

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.

Answer 2

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 `(<0)` value of the derivative is indicative of the function decreasing at such a point.

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

`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 `x=-1`, we must find `f'(-1)`.

`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(x)=-xln(2x^2)`:

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

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