The derivative of a function f is given by f'(x)= (x-3)e^x for x>0 and f(x)=7? a.) The function has a critical point at x=3. At this point, does f have a relative minimum or neither?

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The derivative of a function f is given by f'(x)= (x-3)e^x for x>0 and f(x)=7? a.) The function has a critical point at x=3. At this point, does f have a relative minimum or neither?

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Answer 1 · 2016-12-09T17:13:27

The function $f (x)$ $f(x)$ has a local maximum at $x = 3$ $x=3$ . See explanation.

Explanation:

To find if a function has a critical point at a place where $f'(x)=0$ you have to check if the derivative changes sign at this point. If the change occurs then #f(x) has:

Minimum if $f'(x)$ changes sign from negative to positive
Maximum if $f'(x)$ changes sign from positive to negative.

To check it you can calculate the second derivative:

$f''(x)=1*e^x-(x-3)e^x=(1-x+3)e^x=(2-x)e^x$

$f''(3)=(2-3)e^3=-e^3<0$

$f''(x)$ is negative in $x=3$ . This means that $f'(x)$ is decreasing at $x=3$ , this finally means that $f(x)$ has a MAXIMUM at $x=3$

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The derivative of a function f is given by f'(x)= (x-3)e^x for x>0 and f(x)=7? a.) The function has a critical point at x=3. At this point, does f have a relative minimum or neither?

1 Answer

Explanation:

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