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Mathematics 8 Online
OpenStudy (anonymous):

find the critical points on the given interval \[2^{x}\sin x; [-2,6]\]

OpenStudy (abb0t):

Start by finding the derivative of the function

OpenStudy (abb0t):

By definition, x=c is a critical point of the function if: f"(c) = 0 or f'(c) = DNE

OpenStudy (anonymous):

\[2^{x}\ln 2(\sin x) + 2^{x}\cos x\] right

OpenStudy (abb0t):

Correct.

OpenStudy (abb0t):

Now, apply the definition I just provided. Set your deriviative equal to zero.

OpenStudy (anonymous):

well this is where i'm lost

OpenStudy (abb0t):

Well, you can factor out:\[2^x(\cos(x)+\ln(2)\sin(x))\]\[2^x = 0, \cos(x)+\ln(2)\sin(x)=0\]

OpenStudy (abb0t):

Does that make it a bit easier?

OpenStudy (anonymous):

well i got that far but doesn't that still only leave 2 answers as to what x is

OpenStudy (abb0t):

I think one answer is DNE.

OpenStudy (anonymous):

book claims that it's -.96 , 2.18, 5.32

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