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Mathematics 19 Online
OpenStudy (anonymous):

Find the local maximum value of f using both the First and Second Derivative Tests. f(x) = x + √9 - x

OpenStudy (anonymous):

i already derivate the sqrt9-x to (9-x)^(1/2)

OpenStudy (anonymous):

but got stuck please shoe me steps but step to the answer so that way i can follow up

OpenStudy (tad1):

If you wrote f(x) correctly, x + (-x) = 0 and f(x) become sqrt 9 which is a constant, the derivative would be 0. By firts derivative test that would be a critical point. the 2nd derivative would also be 0, indicating a local max.

OpenStudy (anonymous):

I think he meant to say\[f(x)=x +\sqrt{9-x}\]right?

OpenStudy (anonymous):

\[f'(x)=1-(1/2)(9-x)^{1/2}\]

OpenStudy (anonymous):

yes im up to there now i need to solve it ....

OpenStudy (anonymous):

You set f'(x)=0 and solve for x. Let's see you try that.

OpenStudy (anonymous):

\[1-9/(2\sqrt{x})+0?\]

OpenStudy (anonymous):

i meant =0?

OpenStudy (anonymous):

\[(1/2)(9-x)^{1/2}=1\]

OpenStudy (anonymous):

\[(9-x)^{1/2}=2\]

OpenStudy (anonymous):

9-x=4 x=5

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