Question

Find all the critical numberes of x2^(-x^2)

Answer 1

Do you mean X to the power or 2 to the power? @calc_nofori

Answer 2

\[x2^{-x ^{^{2}}}\]

Answer 3

\(\large \color{black}{\begin{align} f(x)=x\times 2^(-x^2)\hspace{.33em}\\~\\
\end{align}}\)
find \(\large \color{black}{\begin{align} f(x)'\hspace{.33em}\\~\\
\end{align}}\)

Answer 4

i just dont know how. i know the multiplication rule but what confuses me is the derivative of 2^-x^2

Answer 5

this is ur number right

\(\large \color{black}{\begin{align} f(x)=x\times 2^{-x^2}\hspace{.33em}\\~\\
\end{align}}\)

Answer 6

yes

Answer 7

\[\frac{d}{dx}a^x=a^x \times \log_{e}a\]
Apply chain rule

Answer 8

\[2^{-x ^{2}}\log_{e} 2\]

Answer 9

then itd be times -2x?

Answer 10

yes

Answer 11

you may also combine the 2 in -2x and the 2 in 2^(-x^2) by adding the exponents

Answer 12

and then i'd set that to 0 to get the critical numbers?

Answer 13

\(\large \color{black}{\begin{align} f(x)=x\times 2^{-x^2}\hspace{.33em}\\~\\
f'(x)=x\times \left(2^{-x^2}\right)'+x'\times \left(2^{-x^2}\right)\hspace{.33em}\\~\\
f'(x)=x\times \dfrac{d}{d(-x^2)}\left(2^{-x^2}\right) \dfrac{d-x^2}{dx}+\left(2^{-x^2}\right)\hspace{.33em}\\~\\
f'(x)=x\times \log 2\times \left(2^{-x^2}\right) (-2x)+\left(2^{-x^2}\right)\hspace{.33em}\\~\\
\end{align}}\)

Answer 14

yep f'(x)=0 and find the values of x, they will be the points on the curve for which you have local maxima or minima

Answer 15

\(\large \color{black}{\begin{align}
f'(x)=\left(2^{-x^2}\right)(1-x^2 \log 4)\hspace{.33em}\\~\\
\left(2^{-x^2}\right)(1-x^2 \log 4)=0\hspace{.33em}\\~\\
\end{align}}\)

Answer 16

it'd be 0 and then for the second part, i wouldnt know how to do it properly. I know the answer is suppose to be \[1/\sqrt{2\ln2}\]

Answer 17

but i dont get that answer

Answer 18

\(\large \co\(\large \color{black}{\begin{align}
f'(x)&=\left(2^{-x^2}\right)(1-x^2 \log 4)\hspace{.33em}\\~\\
\left(2^{-x^2}\right)(1-x^2 \log 4)&=0\hspace{.33em}\\~\\
\left(2^{-x^2}\right)&=0\hspace{.33em}\\~\\
\text{or}\hspace{.33em}\\~\\
(1-x^2 \log 4)&=0\hspace{.33em}\\~\\
\left(2^{-x^2}\right)&=2^0\hspace{.33em}\\~\\
\text{or}\hspace{.33em}\\~\\
x^2 \log 4&=1\hspace{.33em}\\~\\
-x^2&=0\hspace{.33em}\\~\\
\text{or}\hspace{.33em}\\~\\
x^2& =\dfrac{1}{\log 4}\hspace{.33em}\\~\\
x&=0\hspace{.33em}\\~\\
\text{or}\hspace{.33em}\\~\\
x&=\pm\sqrt{\dfrac{1}{2\log 2}}\hspace{.33em}\\~\\
\end{align}}\)

Answer 19

forget /ignore the red

Answer 20

@mathmath333 wouldn't 2^0 be equal to 1 and not 0

Answer 21

oh yep let me correct

Answer 22

how do you get it to be \[\sqrt{\frac{ 1 }{ 2\log2 }}\]

Answer 23

what do you mean @calc_nofori The solution of the equation is x=1/sqrt(2log2). That's about it, if you want to find the point, you'll have to put the value of x in your f(x), that will give you your 'y'.

Answer 24

\(\large \color{black}{\begin{align}
\left(2^{-x^2}\right)&=0\hspace{.33em}\\~\\
\log \left(2^{-x^2}\right)&=\log 0\hspace{.33em}\\~\\
-x^2\log \left(2\right)&=-\infty \hspace{.33em}\\~\\
-x^2&=-\infty \hspace{.33em}\\~\\
x&=\infty
\end{align}}\)