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

Question

William Abdalla · Accepted Answer

#f(x)# is neither increasing nor decreasing when #x=1#.

Find #f(1)#. If #f(1)>0#, the function is increasing when #x=1#. If #f(1)<0#, the function is increasing when #x=1#. If #f(1)=0#, there is a critical value and the function is neither increasing nor decreasing. To find #f(x)#, use the product rule, which states that for a function #f(x)=g(x)h(x)#, then #f'(x)=g'(x)h(x)+h'(x)g(x)# First, note that I'm assuming you meant #(ln(x))^2#, and not #ln(x^2)#. #f'(x)=(ln(x))^2d/dx(x)+xd/dx((ln(x))^2)# Determine every derivative: #d/dx(x)=1# The following calls for the chain rule: #d/dx((ln(x))^2)=2ln(x)d/dx(ln(x))=2ln(x)(1/x)# Plug these back in to find #f'(x)#. #f'(x)=(ln(x))^2(1)+x(2ln(x)(1/x))# Simplify. #f'(x)=(ln(x))^2+2ln(x)# Now, find #f'(1)#. #f'(1)=(ln(1))^2+2ln(1)=0+0=0# Thus, #f(x)# is neither increasing nor decreasing when #x=1#. This is the graph of #f(x)#: graph{x (ln(x))^2 [-1.46, 5.565, 10.887, -3.16]} There is a minimum when #x=1#.

Isabella Ackerman · Answer

undefined To determine if $ f(x) = x \ln(x)^2 $ is increasing or decreasing at $ x = 1 $, we need to evaluate the derivative of $ f(x) $ at $ x = 1 $ and check its sign.

Taking the derivative of $ f(x) $ with respect to $ x $ using the product rule, we get:

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

Simplify the expression:

$$ f'(x) = \ln(x)^2 + 2 \ln(x) $$

Evaluate $ f'(1) $:

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

Since $ \ln(1) = 0 $, we have:

$$ f'(1) = 0^2 + 2 \cdot 0 = 0 $$

Since $ f'(1) = 0 $, we can't determine if $ f(x) $ is increasing or decreasing at $ x = 1 $ using the first derivative test. Further analysis, such as examining the behavior of the function on both sides of $ x = 1 $, would be needed to determine the behavior of the function at that point.