MHB 1.8.4 AP Calculus Exam Integral of piece-wise function

[karush]

View attachment 9418
image due to macros in Overleaf

ok I think (a) could just be done by observation by just adding up obvious areas

but (b) and (c) are a litte ?

sorry had to post this before the lab closes

 

[skeeter]

Is this the intro to the problem you're trying to cite ?

 

[karush]

Is this the intro to the problem you're trying to cite ?

Yeah I have seen it in several AP books but never did it.

 

[skeeter]

image due to macros in Overleaf

ok I think (a) could just be done by observation by just adding up obvious areas

but (b) and (c) are a litte ?

sorry had to post this before the lab closes

correction ... $\displaystyle g(x) = \int_1^x f(t) \,dt$

(a) $\displaystyle g(2) = \int_1^2 f(t) \,dt = \dfrac{1}{4}$

$\displaystyle g(-2) = \int_1^{-2} f(t) \,dt = -\int_{-2}^1 f(t) \,dt = \dfrac{\pi - 3}{2}$

(b) from the FTC, $g'(x) = f(x) \implies g''(x) = f'(x)$

$g'(3) = f(3) = -1$, $g''(3) = f'(3) = -\dfrac{1}{2}$

(c) horizontal tangents to $g \implies g' = f = 0$

$f = 0$ at $x = \pm 1$ ... relative max at $x=-1$ because $g'$ changes sign from (+) to (-)

neither a max or min at $x=1$ because $g' = f$ does not change sign.

(d) inflection points occur where $g''=f'=0$ or is undefined and $g''=f'$ changes sign.

$g'' = f'$ equals 0 and is undefined at $x = \pm 1$ ... inflection point at $x=1$

$g'' = f'$ is undefined at $x=-2$ ... inflection point there, also because $g''=f'$ changes sign.

 

[karush]

mahalo

That was a great help...

I did look at some other free responses, but it was messy...