Question

What is the derivative of (-x^2)(e^(x/2))+1

Step by step please :)

Answer 1

\[-x^2(e^{\frac{1}{2}}+1)\] is either a trick question or a typo

Answer 2

\(e^{\frac{1}{2}}+1\) is a constant.
is there an \(x\) involved some where in that part?

Answer 3

\[(-x ^{2} * e ^{1/2}) + 1\]

Answer 4

You have the parentheses in the wrong place satellite, the '+1' does not go in there with the e

Answer 5

but it is \(e^{\frac{1}{2}}=\sqrt{e}\) right? not \(e^{\frac{1}{2}x}\) for example

Answer 6

the derivate of 1 is zero
the derivate de polinomio for constante is the derivate of polinomio for te constant
constant=\[-e ^{1/2}\]

Answer 7

then the derivative is \(-2\sqrt{e}x\) and that is all

Answer 8

Right, but the one is just there as a distraction, per se

Answer 9

since \(e^{\frac{1}{2}}\) is just a number

Answer 10

then derivate is \[2*x*(-e ^{1/2}) + 0\]

Answer 11

or \[-2xe ^{1/2}\]

Answer 12

Ankolix are you sure??

Answer 13

yeah! only if the equation that you put was \[x ^{2}*(e ^{1/2})*(-1) + 1\]

Answer 14

Yes that's right. Do you know how to get the critical values? I have x=o as one, but I need the second one

Answer 15

when the derivate is 0... so, i think that only answer is x=0

Answer 16

Well... My Teacher said that I am missing a critical value... I'm so confused :(

Answer 17

can you write the exact original problem?
the only place where \(-2\sqrt{e}x=0\) is if \(x=0\)
i get the feeling something is missing

Answer 18

perhaps it was something like
\[-x^2e^{\frac{x}{2}}\]
not to second guess you

Answer 19

yeah, im agree with satellite73

Answer 20

I promise you that the equation is (-x^2)(e^(1/2)) +1

Answer 21

It's completely possible that he made a mistake when grading my paper

Answer 22

Satellite my mistake, it is e^x/2

Answer 23

Hello?

Answer 24

got logged out

Answer 25

@satellite73 sorry to bother anyone here, if you have a chance could you possibly help me with my problem when you are done here? thanks.

Answer 26

\[f(x)=-x^2e^{\frac{x}{2}}+1\]
you need the product rule for the derivative

Answer 27

\[f'(x)=-2xe^{\frac{x}{2}}-\frac{1}{2}x^2e^{\frac{x}{2}}\]

Answer 28

factor out the \(xe^{\frac{x}{2}}\) and then set equal to zero

Answer 29

\[f'(x)=-xe^{\frac{x}{2}}\left(2+\frac{1}{2}x\right)\]

Answer 30

i see, i see..... ok Thank you this helped a lot :)

Answer 31

one zero is if \(x=0\) the other if \(2+\frac{1}{2}x=0\) etc

Answer 32

yw

Answer 33

@erin i don't see your problem

Answer 34

@satellite73 i'll repost it now