Question

AP Calculus Help Please!

Answer 1

4. By the fundamental theorem of calculus,


`\displaystyle(\mathrm dg(x))/(\mathrm dx)=(\mathrm d)/(\mathrm dx)\int_(-2)^xf(t)\,\mathrm dt=f(x)`


`g(x)` is increasing on those intervals where
`g'(x)=(\mathrm dg(x))/(\mathrm dx)>0`. So you have


`g'(x)=f(x)=\begin{cases}3&\text{for }-3\le x<0\\-x+3&\text{for }0\le x\le6\\-3&\text{for }6<x\le9\end{cases}`

which is clearly positive for
`x\in[-3,0)`, and in the second interval you have


`-x+3>0\implies x<3`

Together, this means
`g'(x)>0` for all
`x\in[-3,3)`.

5. When
`0\le x\le6`,
`f(x)` reduces to
`-x+3`, so you have


`g(x)=\displaystyle\int_(-2)^xf(t)\,\mathrm dt=\int_(-2)^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt`

`g(x)=6+\left(3t-\frac12t^2\right)\bigg|_(t=0)^(t=x)`

`g(x)=6+3x-\frac12x^2`