According to the text book, the correct answer is $(4, \frac{2}{3})$
When I try to work it out, I get $x = \frac{9}{4}$
It would be tedious to put my attempt here and I'm sure this is simple for anyone who has experience with differentiation.
Thanks in advance!
If we compute $y'(x)$ from $y = \frac{1}{3}x^{\frac{3}{2}} - x^{\frac{1}{2}}$ we get: $y'(x) = \frac{1}{2}x^{\frac{1}{2}} -\frac{1}{2}x^{-\frac{1}{2}}$.
We know that $y'(x) = \frac{3}{4}$, thus $\frac{1}{2}x^{\frac{1}{2}} -\frac{1}{2}x^{-\frac{1}{2}} = \frac{3}{4} \Leftrightarrow x=4$.
And $y(4)=\frac{2}{3}$, so the answer is indeed: $(4, \frac{2}{3})$.
Edit (answer):
$$\frac{1}{2}x^{\frac{1}{2}} -\frac{1}{2}x^{-\frac{1}{2}} = \frac{3}{4} \Leftrightarrow x=4$$.
Let $u = x^{\frac{1}{2}}$, then:
$$u -\frac{1}{u} =\frac{3}{2}$$ $$u^2 -\frac{3}{2}u -1 =0$$ $$(u-2)(u+\frac{1}{2})=0$$
Thus: $u=2\lor u=-\frac{1}{2}$
$\Rightarrow x=u^2=4$