How do you find the critical points for #xlnx# and the local max and min?

Answer 1
The derivative of #xlnx# is given by the product rule.
#y' = 1(lnx) + x(1/x)#
#y' = lnx + 1#
The critical points occur when the derivative equals #0# or is undefined (the latter will only be a critical point if the point is defined in the original function).
#0 = lnx + 1#
#-1 = lnx#
#e^-1 = x#
The derivative is undefined at #x = 0#, but the function is as well, so we can't count it as a critical point.
Whenever #x > e^-1#, the derivative is positive, therefore the function is increasing. This signifies that #x = e^-1# is an absolute minimum.

Here is a graphical confirmation.

graph{xlnx [-18.02, 18.01, -9.01, 9.01]}

Hopefully this helps!

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Answer 2
To find the critical points of \( x \ln(x) \) and determine local maximum and minimum points, follow these steps: 1. Compute the derivative of \( x \ln(x) \) using the product rule: \[ f'(x) = \frac{d}{dx}(x) \cdot \ln(x) + x \cdot \frac{d}{dx}(\ln(x)) \] \[ f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} \] \[ f'(x) = \ln(x) + 1 \] 2. Find critical points by setting the derivative equal to zero and solving for \( x \): \[ \ln(x) + 1 = 0 \] \[ \ln(x) = -1 \] \[ x = e^{-1} \] 3. Determine the nature of critical points using the first derivative test or the second derivative test. - First Derivative Test: - If \( f'(x) > 0 \) for \( x < e^{-1} \) and \( f'(x) < 0 \) for \( x > e^{-1} \), the function has a local maximum at \( x = e^{-1} \). - If \( f'(x) < 0 \) for \( x < e^{-1} \) and \( f'(x) > 0 \) for \( x > e^{-1} \), the function has a local minimum at \( x = e^{-1} \). - Second Derivative Test: - Compute the second derivative \( f''(x) \). - If \( f''(e^{-1}) > 0 \), the function has a local minimum at \( x = e^{-1} \). - If \( f''(e^{-1}) < 0 \), the function has a local maximum at \( x = e^{-1} \). 4. Compute \( f''(x) \) and evaluate it at \( x = e^{-1} \): \[ f''(x) = \frac{d}{dx}(\ln(x) + 1) \] \[ f''(x) = \frac{1}{x} \] Evaluate at \( x = e^{-1} \): \[ f''(e^{-1}) = \frac{1}{e^{-1}} = e \] Since \( f''(e^{-1}) > 0 \), the function has a local minimum at \( x = e^{-1} \).
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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.

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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