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Pls help maths urgent

Can somebody please help? I have done my working out - but not sure where I've gone wrong. Will be so much appreciated if somebody could please help me!
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In product rule you’re meant to add the functions not subtract one from the other.
Reply 2
Original post by TheTree0fDeath
In product rule you’re meant to add the functions not subtract one from the other.

Thanks a lot! So i corrected that and got 0= -2x^4e^(-x^2) + 3x^2e^(-x^2)
Not sure how to simplify this though -could you pls kindly help with this as well
have just attached ms below
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(edited 5 years ago)
Original post by h26
Thanks a lot! So i corrected that and got 0= -2x^4e^(-x^2) + 3x^2e^(-x^2)
Not sure how to simplify this though -could you pls kindly help with this as well


Yeah sure, basically you want to divide everything by e^(-x^2) which you’re allowed to do because that function never equals zero.
Reply 4
Original post by TheTree0fDeath
Yeah sure, basically you want to divide everything by e^(-x^2) which you’re allowed to do because that function never equals zero.

I don't quite get that - I thought you never cancel out something which has a variable in it e.g. x in this case
And am not sure about what you said about the function never equaling zero and also how its related.
lemme know please :smile:
Original post by h26
I don't quite get that - I thought you never cancel out something which has a variable in it e.g. x in this case
And am not sure about what you said about the function never equaling zero and also how its related.
lemme know please :smile:


Ok I’m going to have to get into some deep technical stuff here so bear with me xD.

So the reason you never cancel something with a variable is that you lose solutions if it equals zero.

So for x^2 = 7x I cant divide by x because I lose the solution x=0.

For sin^2(x) = 4cos(x)sin(x) I can’t divide by sin(x) because I lose the solution sin(x)=0.

However in the case of e^(-x^-2) this function can never be zero so we are allowed to cancel it.

If you’re not convinced then factorise it by taking out a factor of e^(x^-2) and then solve what’s in the brackets. Obviously if the bit inside the brackets = 0 then that’s a solution. The bit outside the brackets has no solutions because e^(-x^2) can never be zero.
(edited 5 years ago)
Reply 6
Original post by TheTree0fDeath
Ok I’m going to have to get into some deep technical stuff here so bear with me xD.

So the reason you never cancel something with a variable is that you lose solutions if it equals zero.

So for x^2 = 7x I cant divide by x because I lose the solution x=0.

For sin^2(x) = 4cos(x)sin(x) I can’t divide by sin(x) because I lose the solution sin(x)=0.

However in the case of e^(-x^-2) this function can never be zero so we are allowed to cancel it.

If you’re not convinced then factorise it by taking out a factor of e^(x^-2) and then solve what’s in the brackets. Obviously if the bit inside the brackets = 0 then that’s a solution. The bit outside the brackets has no solutions because e^(-x^2) can never be zero.

Ah thanks a lot -never knew that!:smile: Ik it's pretty late so nw if not possible but if you do get the chance..how do you do 9 iv A -it's from the same question
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Thanks a lot again:smile:
Original post by h26
Ah thanks a lot -never knew that!:smile: Ik it's pretty late so nw if not possible but if you do get the chance..how do you do 9 iv A -it's from the same question
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Thanks a lot again:smile:


From t = x^2, we get dt/dx = 2x, so dt = 2x dx, so dx = 1/(2x) dt. Thus we get integral of x^3*e^(-x^2)*1/(2x) dt = 1/2*x^2*e^(-x^2) dt = 1/2*t*e^(-t) dt.

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