1
$\begingroup$

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

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 $f'(t) = 0$.

$t = 3/4$

  • 0
    You've already solved it then?2012-10-22
  • 0
    So what is the actual answer, then? Would I just say that f(t) = 3/4? Is that what the question is asking?2012-10-22
  • 1
    What should $t$ be if $f'(t) = 0$? (see b) And what is the function value of $f$ for that value of $t$?2012-10-22
  • 0
    @TMM: Thanks, that helped me understand what the question was asking for.2012-10-22

2 Answers 2

1

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

2

You already solved it:

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

  • 0
    @Thomas: But let him think what should he do next?2012-10-22
  • 1
    @BabakSorouh: Ok2012-10-22