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What are the critical values, if any, of #f(x)=xe^(-3x)#?

Answer 1

Luke Alvarez

#P(1/3, 1/(3e))#

First we derivate this, so

#y = xe^(-3x)#

#ln(y) = ln(x) -3x#

#1/y*dy/dx = 1/x - 3#

#dy/dx = e^(-3x) -3xe^(-3x)#

Critical points are those where the derivative is 0, so

#0 = e^(-3x) - 3xe^(-3x)#

#3xe^(-3x) = e^(-3x)#

Because the exponential is never 0 you can divide it out

#3x = 1# #x = 1/3#

The function has a critical point, in #P(1/3, 1/(3e))#

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Answer 2

Luke Alvarez

To find the critical values of ( f(x) = xe^{-3x} ), we first need to find its derivative, ( f'(x) ), and then solve for ( x ) when ( f'(x) = 0 ).

[ f'(x) = (xe^{-3x})' = (x)'e^{-3x} + x(e^{-3x})' = e^{-3x} - 3xe^{-3x} ]

Now, set ( f'(x) = 0 ) and solve for ( x ):

[ e^{-3x} - 3xe^{-3x} = 0 ]

Factoring out ( e^{-3x} ), we get:

[ e^{-3x}(1 - 3x) = 0 ]

This equation equals zero when either ( e^{-3x} = 0 ) or ( 1 - 3x = 0 ). However, ( e^{-3x} ) can never be zero, so we solve ( 1 - 3x = 0 ) instead:

[ 1 - 3x = 0 \implies x = \frac{1}{3} ]

Therefore, the critical value of ( f(x) = xe^{-3x} ) is ( x = \frac{1}{3} ).

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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.