Expert: Paul Klarreich - 1/31/2010

Question
1) Where is the graph of f(x) = e^(-x^2) concave down?



2) Find an equation of the line tangent to the curve:

y^2 + 4e^y = 4x + ln(x^2)



Thanks so much!



Carolyn

Answer
Questioner: Carolyn
Country: United States
Category: Calculus
Question: 1) Where is the graph of f(x) = e^(-x^2) concave down?

It will be concave down when  f''(x) is negative.

Find f' = -2x e^(-x^2), then

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

f''(x)  =  e^(-x^2)[(-2x)(-2x ) + (-2))

f''(x)  =  e^(-x^2)[4x^2 - 2]

Now set  that = 0 to find your inflection points and you are on your way.

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2) Find an equation of the line tangent to the curve:

y^2 + 4e^y = 4x + ln(x^2)

>> There are a lot of them.  How about a point?