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

`(d)/(dx) √(x+3) = (1)/(2√(x+3))`

So, the integral is almost immediate:

`\displaystyle\int (2)/(√(x+3)) = \int (4)/(2√(x+3)) = 4\int(1)/(2√(x+3)) = 4√(x+3)+c`

So, up to some constant additive term, our function is
`4√(x+3)+c`

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

`f(6) = 4√(6+3)+c = 4√(9)+c=12+c=10 \iff c=-2`

And so our function is
`4√(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√(x+3)-2=6 \iff 4√(x+3) = 8 \iff √(x+3) = 2 \iff x+3=4 \iff x = 1`