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$f(t) = 8t^{1/2} + 6t^{-1/2}$

Somehow I think the question is related to the previous parts, where I did:

a) Find an expression for $f'(t)$.

$f'(t) = 4t^{-1/2} - 3t^{-3/2}$

b) Find the value for $t$ for which $f'(t) = 0$.

$t = 3/4$

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    @TMM: Thanks, that helped me understand what the question was asking for.2012-10-22

2 Answers 2

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The only thing left is to find $ f\left(\frac{3}{4}\right) = 8\left(\frac{3}{4}\right)^{1/2} + 6\left(\frac{3}{4}\right)^{-1/2} = \dots $

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You already solved it:

$f'(t)=\frac{4}{t^{1/2}}-\frac{3}{t\cdot t^{1/2}}=0\Longleftrightarrow 4t-3=0\,\,,\,\,so...$

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    @BabakSorouh: Ok2012-10-22