Home > Calculus > Identifying Stationary Points (Critical Points) for a Function

What are the critical values, if any, of #f(x) = xe^(2x)#?

Answer 1

Ezra Adams

Critical Value: #x=-1/2#

#f(x)=xe^(2x)#

By Product Rule,

#f'(x)=1 cdot e^(2x)+x cdot 2e^(2x)=(1+2x)e^(2x)#

By setting #f'(x)=0#,

#(1+2x)e^(2x)=0#

By dividing both sides by #e^(2x)#,

#1+2x=0#

By subtracting 1 from both sides,

#2x=-1#

By dividing both sides by #2#,

#x=-1/2# (Critical Value)

I hope that this was clear.

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Answer 2

Emma Aldridge

To find the critical values of ( f(x) = xe^{2x} ), we first need to find its derivative.

The derivative of ( f(x) ) with respect to ( x ) can be found using the product rule:

[ f'(x) = \frac{d}{dx}(x) \cdot e^{2x} + x \cdot \frac{d}{dx}(e^{2x}) ]

[ f'(x) = 1 \cdot e^{2x} + x \cdot 2e^{2x} ]

[ f'(x) = e^{2x} + 2xe^{2x} ]

To find the critical values, we set the derivative equal to zero and solve for ( x ):

[ e^{2x} + 2xe^{2x} = 0 ]

[ e^{2x}(1 + 2x) = 0 ]

This equation is equal to zero if either ( e^{2x} = 0 ) or ( 1 + 2x = 0 ). However, ( e^{2x} ) can never equal zero for any real value of ( x ), so we only need to solve ( 1 + 2x = 0 ):

[ 1 + 2x = 0 ]

[ 2x = -1 ]

[ x = -\frac{1}{2} ]

Therefore, the critical value of ( f(x) = xe^{2x} ) is ( x = -\frac{1}{2} ).

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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.