If y=e−x2, what are the points of inflection, concavity and critical points?

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eslasadanv3 · Accepted Answer

Step 1Given: f(x)=e−x2we can calculate the first and second order derivatives:f′(x)=−2xe−x2fx)=−2e−x2+4x2e−x2=2(2x2−1)e−x2we can therefore determine that:(1) By solving the equation:f′(x)=0⇒−2xe−x2=0we can see that f(x) has a single critical point for x=0, this point is a relative maximum since f0)=−2&amp;lt;0Looking at the second derivative, we can see that 2e−x2 is always positive and non null, so that inflection points and concavity are determined by the factor (2x2−1).Step 2(2) f(x) has two inflection points for:2x2−1=0⇒x=±12Step 3(3) As (2x2−1) is a second order polynomial with leading positive coefficient, we know that is is negative in the inteval between the roots, and positive outside, so:f(x) is concave up in (−∞,−12) and in (12,+∞)f(x) is concave down in (−12,12)

If y=e^{-x^{2}}, what are the points of inflection, concavity and

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