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

Question

Cameron Alexander · Accepted Answer

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

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.

William Abdalla · Answer

Decreasing.

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.

Aurora Alvarado · Answer

undefined To determine whether the function $ f(x) = -x \ln(2x^2) $ is increasing or decreasing at $ x = -1 $, we need to examine the sign of the derivative of the function at that point.

To find the derivative of $ f(x) $, we'll use the product rule and the chain rule.

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

Evaluating this derivative at $ x = -1 $, we have:

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

Since $ \ln(2) $ is positive, $ f'(-1) $ will be negative, and hence, the function $ f(x) $ is decreasing at $ x = -1 $.