# Thread: Concavity and the Derivative of e^x

1. ## Concavity and the Derivative of e^x

Let f be the function given by f(x)=2xe^(x). The graph of f is concave down when A)x<-2 B)x>-2 C)x<-1 D)x>-1 E)x<0

I know that you can use the 2nd derivative to find concavity. If f''(x) is negative, it is concave down and if its f''(x) is positive, its concave up

However, i don't know how to find f''(x) of this function

2. Originally Posted by rawkstar
Let f be the function given by f(x)=2xe^(x). The graph of f is concave down when A)x<-2 B)x>-2 C)x<-1 D)x>-1 E)x<0

I know that you can use the 2nd derivative to find concavity. If f''(x) is negative, it is concave down and if its f''(x) is positive, its concave up

However, i don't know how to find f''(x) of this function
$f(x) = 2x \cdot e^x$

product rule ...

$f'(x) = 2x \cdot e^x + 2e^x$

$f'(x) = 2e^x(x + 1)$

do the product rule again ...

3. i'm guessing the the derivative of 2e^x is still 2e^x

is it

f''(x)=(2e^x)[1+(x+1)]

4. Originally Posted by rawkstar
i'm guessing the the derivative of 2e^x is still 2e^x

is it

f''(x)=(2e^x)[1+(x+1)]
simplify ...

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

now finish the problem

5. ok so
0=(2e^x)(x+2)
I distributed to get
0=2xe^x+4e^x
and i don't know how to solve from here (i'm not quite sure how to use natural log)

6. $f''(x) = 2e^x(x+2) = 0$

$x = -2$