What are the local extrema of #f(x)= xlnx-xe^x#?

Answer 1

This function has no local extrema.

#f(x)= xlnx-xe^x implies#
#g(x) equiv f^'(x) = 1+lnx - (x+1)e^x#
For #x# to be be a local extremum, #g(x)# must be zero. We will now show that this does not occur for any real value of #x#.

Note that

#g^'(x) = 1/x-(x+2)e^x,qquad g^{''}(x) = -1/x^2-(x+3)e^x#
Thus #g^'(x)# will vanish if
#e^x = 1/(x(x+2))#
This is a transcendental equation which can be solved numerically. Since #g^'(0) = +oo# and #g^'(1)=1-3e<0#, the root lies between 0 and 1. And since #g^{''}(0) <0# for all positive #x#, this is the only root and it corresponds to a maximum for #g(x)#
It is quite easy to solve the equation numerically, and this shows that #g(x)# has a maximum at #x=0.3152# and the maximum value is #g(0.3152) = -1.957#. Since the maximum value of #g(x)# is negative, there is no value of #x# at which #g(x)# vanishes.

It may be instructive to look at this graphically:

graph{xlog(x)-xe^x [-0.105, 1, -1.175, 0.075]}

As you can see from the graph above, the function #f(x)# actually has a maximum at #x=0# - but this is not a local maximum. The graph below shows that #g(x)equiv f^'(x)# never takes the value zero.

graph{1+log(x)-(x+1)*e^x [-0.105, 1, -3, 0.075]}

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the local extrema of ( f(x) = x \ln(x) - xe^x ), we need to find the critical points by setting the derivative equal to zero and then checking the second derivative to determine whether each critical point corresponds to a local minimum, local maximum, or neither.

  1. Find the first derivative: [ f'(x) = \ln(x) + 1 - e^x - xe^x ]

  2. Set ( f'(x) ) equal to zero and solve for ( x ) to find critical points.

  3. Once you have critical points, evaluate the second derivative at each critical point to determine the concavity and thus the nature of the extrema.

  4. If ( f''(x) > 0 ), it's a local minimum. If ( f''(x) < 0 ), it's a local maximum. If ( f''(x) = 0 ), the test is inconclusive.

  5. Any critical points that yield a change in sign of the second derivative indicate a local extremum.

  6. Additionally, check the behavior of the function at the boundaries of the domain (if any) to ensure no additional extrema are missed.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7