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

2 Answers
Apr 4, 2016

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.

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 #(<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.