Question

Please solve i and ii for me

Answer 1

If you know the derivative $f'(x)$ of some function $f(x)$, you can tell exactly who $f(x)$ is, up to an additive, constant term. In fact, knowing $f'(x)$, you have

$\displaystyle \int f'(x) = f(x)+c$

In your case, we have

$\dfrac{d}{dx} \sqrt{x+3} = \dfrac{1}{2\sqrt{x+3}}$

So, the integral is almost immediate:

$\displaystyle\int \dfrac{2}{\sqrt{x+3}} = \int \dfrac{4}{2\sqrt{x+3}} = 4\int\dfrac{1}{2\sqrt{x+3}} = 4\sqrt{x+3}+c$

So, up to some constant additive term, our function is $4\sqrt{x+3}+c$

To fix this constant, we know that the function passes through the point (6,10), so we have

$f(6) = 4\sqrt{6+3}+c = 4\sqrt{9}+c=12+c=10 \iff c=-2$

And so our function is $4\sqrt{x+3}-2$

If we want to know when this function equals 6, we simply have to ask $f(x)=6$ and solve for x, so we have

$4\sqrt{x+3}-2=6 \iff 4\sqrt{x+3} = 8 \iff \sqrt{x+3} = 2 \iff x+3=4 \iff x = 1$