Concavity and Inflection Point

[winterrose]

Hello so stuck here, could use some help.


Given f(x)=(x^4)(e^-x)

Step 1) Find f'(x)

• f'(x)=x^4+(4x)e^-x

Step 2) Find f"(x), this is where I am having trouble

• f"(x)=(e^-x)(x^4-4x^3-4x+4)

Step 3) Find f"(x)=0


So I set e^-x=0 and x^4-4x^3-4x+4=0, to find where f"(x)=0


e^-x=0
x=0


x^4-4x^3-4x+4=0
(I am at a lost on how to solve this part)


My first attempt is [x^3(x-4)-4(x+1)] not sure if that does anything. My algebra is a little rusty, and it is probably something silly but I could use any suggestion, cheers.

 

[Romsek]

Hello so stuck here, could use some help.


Given f(x)=(x^4)(e^-x)

Step 1) Find f'(x)

f'(x)=x^4+(4x)e^-x
​

Ok this is incorrect for a start.

Given
\(\displaystyle f(x)=x^4 e^{-x}\)
\(\displaystyle f'(x)=e^{-x} \left(4 x^3-x^4\right)\)

Start from here and see if you can work it through. You have the right idea.

 

[winterrose]

Ok this is incorrect for a start.

Given
\(\displaystyle f(x)=x^4 e^{-x}\)
\(\displaystyle f'(x)=e^{-x} \left(4 x^3-x^4\right)\)

Start from here and see if you can work it through. You have the right idea.

Thanks yep that was the problem right there, going too fast and missed that simple mistake. Thanks a lot!!