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[SOLVED] 4.2.251 AP calculus exam concave ?

karush

Well-known member
Jan 31, 2012
2,678
Capture.PNG

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

Well-known member
MHB Math Helper
Mar 1, 2012
655
$g’ > 0 \text{ for all } x \in (0,2) \implies g \text{ is increasing}$

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

karush

Well-known member
Jan 31, 2012
2,678
Capture.PNG

how???
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655
$\displaystyle g'(x) = \int_0^x e^{-t^2} \, dt$ is the area accumulation function shown in the graph.

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

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