Question

Click here please. The problems are in picture format

Answer 1

First picture:

You got the first two right (good job!), so I'll just tackle the third one. If you have to compute
`f(x-4)`, it means that you have to substitute
`x-4` in place of
`x`. Differently from the first two points, this will generate a new function, rather than a specific value:

`f(x) = √(x+4)+2 \implies f(x-4) = √((x-4)+4)+2 = √(x)+2`

Second picture:

Given the point
`(x,y)` highlighted in the picture, you can deduce that the base is
`2x` units long (since it spans from
`(-x,0)` to
`(x,0)`) and the height is
`y` units long (because it spans from
`(x,0)` to
`(x,y)`). So, the area of the rectangle is the multiplication between base and height:

`A = 2xy`

But we know that
`y=f(x)=√(36-x^2)`, so we have

`A = 2x√(36-x^2)`

The domain of this function is given by the domain of the square root: we want its argument to be non, negative, so we have

`36-x^2 \geq 0 \iff x^2 \leq 36 \iff -6 \leq x \leq 6`

But since the problem is symmetric, the answer is

`0 \leq x \leq 6`

You can only see the answer
`0 < x < 6` because, if you choose
`x=0` or
`x=6`, the rectangle degenerates to a segment, and your exercise doesn't like this scenario, apparently