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:

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 ]

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} ]

Since ( f''(1) ) is negative, the critical point ( x = 1 ) corresponds to a maximum value.

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 ]

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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