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Over which interval does the function decrease the fastest?
f(x)=3√ (-x+2)
A. 3
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@mathmath333
that is where the derivitive has the lowest value \[f(x)=3 \sqrt{-x+2} = 3(-x+2)^\frac{ 1 }{ 2 }\] so \[f'(x)= 1.5(-1)(-x+2)^{-\frac{ 1 }{ 2 }} =-\frac{1.5}{\sqrt{ -x+2}}\] The domain of \(f(x)\) is \((\infty, 2]\) and \[\lim_{x \rightarrow 2} f'(x)= -\frac{1.5}{\sqrt{ -x+2}}\rightarrow -\infty\] so the interval that is closest to -2 and within the domain will be the one we are looking for A and D are not fully in the domain From B and C, B is closest to 2 So it must be B
Thank you @abtster
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