Let F be the function given by f(x)= 2xe^x. Determine the interval(s) on which the graph is concave down.

Hello,

y=2xe^x
y'=2(e^x+xe^x)=2(x+1)e^x
y''=2(e^x+(x+1)e^x)=2(x+2)e^x

x |-infinite -2 0 +infinite
e^x | ++++++++++++++++++
x+2 |------------0 +++++++++++
y'' | -----------0 +++++++++++

y''<0 if x<-2
The interval on which the graph is concave down is (-infinite -2[

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joaobezerra joaobezerra · Answer 2 · 2021-11-12T13:24:10+01:00

Using the second derivative, it is found that the graph is concave down on the interval [tex](-\infty, -2)[/tex].

A function is concave down when the second derivative is negative.

The function is:

[tex]f(x) = 2xe^x[/tex]

The first derivative is as follows, applying the product rule:

[tex]f^{\prime}(x) = 2e^x + 2xe^x[/tex]

The second derivative is the derivative of the first derivative, given by:

[tex]f^{\prime\prime}(x) = 2e^x + 2e^x + 2xe^{x} = 4e^x + 2xe^x = e^x(4 + 2x)[/tex]

The exponential is always positive, so the second derivative is negative if:

[tex]4 + 2x < 0[/tex]

[tex]2x < -4[/tex]

[tex]x < -\frac{4}{2}[/tex]

[tex]x < -2[/tex]

Hence, the graph is concave down on the interval [tex](-\infty, -2)[/tex].

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