Dec 24, 2019 Thread starter #1 karush Well-known member Jan 31, 2012 2,678 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
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
Dec 24, 2019 #2 skeeter Well-known member MHB Math Helper Mar 1, 2012 655 $g’ > 0 \text{ for all } x \in (0,2) \implies g \text{ is increasing}$ $g’’ = e^{-x^2} > 0 \text{ for all } x \in (0,2) \implies g \text{ is concave up}$
$g’ > 0 \text{ for all } x \in (0,2) \implies g \text{ is increasing}$ $g’’ = e^{-x^2} > 0 \text{ for all } x \in (0,2) \implies g \text{ is concave up}$
Dec 24, 2019 #4 skeeter Well-known member MHB Math Helper Mar 1, 2012 655 karush said: how??? Click to expand... how what?
Dec 26, 2019 #5 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. \(\displaystyle \color{red}{g'(0.5) = \int_0^{0.5} e^{-t^2} \, dt} < \color{blue}{g'(1) = \int_0^1 e^{-t^2} \, dt} < \color{green}{g'(1.5) = \int_0^{1.5} e^{-t^2} \, dt}\) Attachments Accumulation function.jpg 132.7 KB Views: 8
$\displaystyle g'(x) = \int_0^x e^{-t^2} \, dt$ is the area accumulation function shown in the graph. \(\displaystyle \color{red}{g'(0.5) = \int_0^{0.5} e^{-t^2} \, dt} < \color{blue}{g'(1) = \int_0^1 e^{-t^2} \, dt} < \color{green}{g'(1.5) = \int_0^{1.5} e^{-t^2} \, dt}\)