4.2.251 AP calculus exam concave ?

[karush]

View attachment 9441

ok this one baffled me a little
but isn't $g = e^{-t^2}$and the graph of that has an inflection point at x=1

 

[skeeter]

$g’ > 0 \text{ for all } x \in (0,2) \implies g \text{ is increasing}$

$g’’ = e^{-x^2} > 0 \text{ for all } x \in (0,2) \implies g \text{ is concave up}$

 

[karush]

View attachment 9442

how?

 

[skeeter]

how?

how what?

 

[skeeter]

$\displaystyle g'(x) = \int_0^x e^{-t^2} \, dt$ is the area accumulation function shown in the graph.

\(\displaystyle \color{red}{g'(0.5) = \int_0^{0.5} e^{-t^2} \, dt} < \color{blue}{g'(1) = \int_0^1 e^{-t^2} \, dt} < \color{green}{g'(1.5) = \int_0^{1.5} e^{-t^2} \, dt}\)