Value(s) of x where the value of y=xLn(2x) is an extremum. Round x to 4 decimal places. State whether the extremum is max or min.

Question

I'm confused on how to set this up. Professor had said to take the inverse function of ln(x) 

which would be e^x but what do I do with that 2 or that extra x?

Answer 1

only way I know is to use calculus. Take the derivative dy/dx and find x values where derivative is zero or undefined. then test any answers in the domain of the function to see if corresponds to a minimum, maximum, or neither.

 

we use the product rule to find the first derivative

 

dy/dx = x (2/2x) + ln(2x) = 1 + ln(2x)

 

set 1 + ln(2x) = 0 and solve for x

 

ln(2x) = -1         now raise e to the powers on both sides

 

e^ln(2x) = e^-1

 

2x = 1/e

 

x = 1/2e = .1839

 

this is a minimum because the second derivative is positive at x = 1/2e which means the function is concave upward

 

the first derivative is undefined at x = 0 so this could be an extremum however x is not in the domain so the possibility is eliminated from consideration

Answer 2

To find the extremum, the derivative of f(x)  = x ln(2x) must be calculated and set to zero.

The derivative f'  can be found with the aid of the chain rule.   It is

f' = 1 + ln(2) + ln(x)           Setting this to zero and using the fact that exp is the inverse of ln  gives

 

x = exp(  -(ln(2) +1)   )      for the value of x at the extreme point.

 

Evaluation via calculator gives   x =  .1839

 

To determine max or min , evaluate the second derivative

 

f'' = 1/x         This is positive for x = .1839   so this is a minimum.

 

This function has another critical point at x = 0.