How do you find the local extremas for #f(x)=xe^x#?
There is a relative minimum at the point
If the derivative is positive, we know that the function is increasing, whereas if the derivative is negative, then the function is decreasing.
When the derivative changes from negative to positive, the function has a local minimum, whereas if the change of sign is reversed, that is, from positive to negative, then the function has a local maximum.
Equaling to zero we have:
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To find the local extrema of the function ( f(x) = xe^x ), you first need to find its critical points by taking the derivative and setting it equal to zero. Then, you can use the second derivative test to determine whether these critical points correspond to local maxima, local minima, or points of inflection.

Take the derivative of ( f(x) ): [ f'(x) = e^x + xe^x ]

Set the derivative equal to zero and solve for ( x ) to find critical points: [ e^x + xe^x = 0 ] [ e^x(1 + x) = 0 ]
This equation yields a critical point at ( x = 1 ).
 Apply the second derivative test: [ f''(x) = e^x + xe^x ]
Evaluate ( f''(1) ): [ f''(1) = e^{1}  e^{1} = 0 ]
Since ( f''(1) = 0 ), the second derivative test is inconclusive. In this case, you can examine the behavior of the function around the critical point. By observing the behavior of the function, you can determine that ( f(x) ) has a local minimum at ( x = 1 ).
Therefore, the local minimum of ( f(x) = xe^x ) occurs at ( x = 1 ).
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To find the local extrema for (f(x) = x e^x), we first find the critical points by setting the derivative equal to zero and solving for (x). Then, we use the second derivative test to determine whether each critical point corresponds to a local maximum, local minimum, or neither.

Find the derivative of (f(x)): [f'(x) = e^x + xe^x]

Set the derivative equal to zero and solve for (x): [e^x + xe^x = 0] [e^x(1 + x) = 0]
This equation equals zero when (e^x = 0) or (1 + x = 0). However, (e^x) is never zero, so we solve (1 + x = 0) for (x): [1 + x = 0] [x = 1]
 Test the critical point (x = 1) using the second derivative test:
Calculate the second derivative of (f(x)): [f''(x) = e^x + e^x + xe^x] [f''(x) = 2e^x + xe^x]
Evaluate (f''(1)): [f''(1) = 2e^{1} + (1)e^{1} = \frac{2}{e}  \frac{1}{e} = \frac{1}{e}]
Since (f''(1) = \frac{1}{e} > 0), the critical point (x = 1) corresponds to a local minimum.
Therefore, the function (f(x) = xe^x) has a local minimum at (x = 1).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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