$f(t) = 8t^{1/2} + 6t^{-1/2}$
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 $f'(t) = 0$.
$t = 3/4$
$f(t) = 8t^{1/2} + 6t^{-1/2}$
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 $f'(t) = 0$.
$t = 3/4$
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 $
You already solved it:
$f'(t)=\frac{4}{t^{1/2}}-\frac{3}{t\cdot t^{1/2}}=0\Longleftrightarrow 4t-3=0\,\,,\,\,so...$