If you know the derivative [tex] f'(x) [/tex] of some function [tex] f(x) [/tex], you can tell exactly who [tex] f(x) [/tex] is, up to an additive, constant term. In fact, knowing [tex] f'(x) [/tex], you have
[tex]\displaystyle \int f'(x) = f(x)+c [/tex]
In your case, we have
[tex] \dfrac{d}{dx} \sqrt{x+3} = \dfrac{1}{2\sqrt{x+3}} [/tex]
So, the integral is almost immediate:
[tex] \displaystyle\int \dfrac{2}{\sqrt{x+3}} = \int \dfrac{4}{2\sqrt{x+3}} = 4\int\dfrac{1}{2\sqrt{x+3}} = 4\sqrt{x+3}+c [/tex]
So, up to some constant additive term, our function is [tex] 4\sqrt{x+3}+c [/tex]
To fix this constant, we know that the function passes through the point (6,10), so we have
[tex] f(6) = 4\sqrt{6+3}+c = 4\sqrt{9}+c=12+c=10 \iff c=-2 [/tex]
And so our function is [tex] 4\sqrt{x+3}-2 [/tex]
If we want to know when this function equals 6, we simply have to ask [tex] f(x)=6[/tex] and solve for x, so we have
[tex] 4\sqrt{x+3}-2=6 \iff 4\sqrt{x+3} = 8 \iff \sqrt{x+3} = 2 \iff x+3=4 \iff x = 1 [/tex]
