# What are the critical points of # f(x) = e^xlnx^2#?

The function

Simplify the expression noting that based on properties of logarithms:

To identify the critical points then we have to solve the equation:

However given:

Apply l'Hospital's:

while:

Consider now:

while:

graph{e^x ln(x^2) [-10, 10, -5, 5]}

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To find the critical points of ( f(x) = e^x \ln(x^2) ), we first need to compute its derivative and then find where the derivative equals zero or is undefined.

Given that ( f(x) = e^x \ln(x^2) ), the derivative of ( f(x) ) with respect to ( x ), denoted as ( f'(x) ), can be found using the product rule of differentiation:

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

Applying the chain rule and derivative of natural logarithm, we have:

[ f'(x) = \left(e^x \cdot \frac{2x}{x^2}\right) + \left(\ln(x^2) \cdot e^x\right) ]

Simplify the expression:

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

[ f'(x) = 2e^x + 2x\ln(x)e^x ]

To find critical points, we set ( f'(x) ) equal to zero and solve for ( x ):

[ 2e^x + 2x\ln(x)e^x = 0 ]

[ 2e^x(1 + x\ln(x)) = 0 ]

The only solution to this equation is when ( 1 + x\ln(x) = 0 ), since ( e^x ) is never zero. Solving this equation for ( x ) would involve numerical methods or graphical analysis.

Therefore, the critical points of ( f(x) = e^x \ln(x^2) ) occur at the solutions to ( 1 + x\ln(x) = 0 ).

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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.

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