How do you determine the maximum and minimum value of the function #f(x)=-xe^x + 2#?
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To find the maximum and minimum values of the function ( f(x) = -xe^x + 2 ), follow these steps:
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Find the critical points by setting the derivative of the function equal to zero and solving for ( x ). [ f'(x) = -e^x - xe^x = 0 ] [ -e^x(1 + x) = 0 ] [ e^x = 0 \quad \text{(No solution)} ] [ 1 + x = 0 ] [ x = -1 ]
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Determine the nature of the critical point using the second derivative test. [ f''(x) = -e^x - e^x - xe^x = -2e^x - xe^x ] [ f''(-1) = -2e^{-1} + e^{-1} = -e^{-1} ]
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Since ( f''(-1) ) is negative, the critical point ( x = -1 ) corresponds to a maximum value.
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Evaluate the function at the critical point and at the endpoints of the interval of interest. [ f(-1) = -(-1)e^{-1} + 2 = e^{-1} + 2 \approx 2.72 ] [ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (-xe^x + 2) = 2 ] [ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} (-xe^x + 2) = -\infty ]
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Compare the values obtained to determine the maximum and minimum.
- Maximum value: ( f(-1) = e^{-1} + 2 \approx 2.72 )
- Minimum value: ( \lim_{x \to -\infty} f(x) = 2 )
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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.
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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