How do you use the first and second derivatives to sketch $f (x) = e^{- x^{2}}$ $f(x)=e^(-x^2)$ ?

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How do you use the first and second derivatives to sketch $f (x) = e^{- x^{2}}$ $f(x)=e^(-x^2)$ ?

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Answer 1 · 2016-12-12T09:18:33

$f (x)$ $f(x)$ has a maximum in $x = 0$ $x=0$ , is strictly increasing for $x < 0$ $x<0$ and strictly decreasing for $x > 0$ $x>0$ , is concave down in the interval $(- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ $(-1/sqrt(2), 1/sqrt(2))$ , concave up outside the interval and has two inflection points for $x = \pm \frac{1}{\sqrt{2}}$ $x=+-1/sqrt(2)$

Explanation:

Using the chain rule we can calculate:

$f (x) = e^{- x^{2}}$ $f(x) = e^(-x^2)$

$f'(x) = -2xe^(-x^2)$

$f''(x) = (-2+4x^2)e^(-x^2)$

So we can see that $f(x)$ is strictly increasing for $x<0$ and strictly decreasing for $x>0$ and hat in $x=0$ it has a critical point.

The second derivative is positive for:

$(4x^2-2)>0$ or $|x|>1/sqrt(2)$

Considering the sign of the second derivative this critical point is a maximum, and $f(x)$ is concave down in the interval $(-1/sqrt(2), 1/sqrt(2))$ , concave up outside the interval and has two inflection points for $x=+-1/sqrt(2)$

graph{e^-(x^2) [-1.293, 1.207, -0.15, 1.1]}

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How do you use the first and second derivatives to sketch $f (x) = e^{- x^{2}}$ $f(x)=e^(-x^2)$ ?

1 Answer

Explanation:

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How do you use the first and second derivatives to sketch f(x)=e^(-x^2)f(x)=e−x2f(x)=e^(-x^2)?

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How do you use the first and second derivatives to sketch $f (x) = e^{- x^{2}}$ $f(x)=e^(-x^2)$ ?