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# C3 help watch

1. Q5 B

Using product rule I got: xe^-x(2-x)

So x=2 and it's a maximum

In the answers it say there is also a minimum at (0,0). I don't know where they got x=0 from?
Attached Images

2. (Original post by Vorsah)
Q5 B

Using product rule I got: xe^-x(2-x)

So x=2 and it's a maximum

In the answers it say there is also a minimum at (0,0). I don't know where they got x=0 from?
dy/dx = 0 <=> x(2-x)e^(-x) = 0 <=> x = 0 or 2-x = 0
3. (Original post by Vorsah)
Q5 B

Using product rule I got: xe^-x(2-x)

So x=2 and it's a maximum

In the answers it say there is also a minimum at (0,0). I don't know where they got x=0 from?
Note that, since

you have

which has TWO solutions
4. (Original post by Vorsah)
Q5 B

Using product rule I got: xe^-x(2-x)

So x=2 and it's a maximum

In the answers it say there is also a minimum at (0,0). I don't know where they got x=0 from?
Stationary point when dy/dx = 0. When x=0 dy/dx=0 therefore there is a stationary point at x=0
5. Thanks

But for Q5 A in the attachment I got e^(-2x)(1-2x) using product rule

So shouldn't it be X=0 or X=0.5

Because in the answers it only says max at 0.5?
6. (Original post by Vorsah)
Thanks

But for Q5 A in the attachment I got e^(-2x)(1-2x) using product rule

So shouldn't it be X=0 or X=0.5

Because in the answers it only says max at 0.5?
No because when x=0 dy/dy=1
7. I need help with Q6 as well.

I got (3x^2)lnx + x^2

= x^2(3lnx + 1)

Don't know what to do from here
No because when x=0 dy/dy=1
So once you get the2 values of X do u have to put them into the gradient function to see if they = 0

Eg for Q5 A I got x=0 or x=0.5 , so do I sub in these values into gradient function and see which one = 0 and then use the value that=0 and ignore the one that doesn't = 0?

When x=0, dy/dx=1

When x = 0.5 dy/dx=0 so this is the correct value
9. (Original post by Vorsah)
I need help with Q6 as well.

I got (3x^2)lnx + x^2

= x^2(3lnx + 1)

Don't know what to do from here
You need to differentiate it and set that equal to 0 to get the stationary points, and then differentiate again to find which is a maximum/minimum.
10. (Original post by The Shanus)
You need to differentiate it and set that equal to 0 to get the stationary points, and then differentiate again to find which is a maximum/minimum.
I have differentiated and got (x^2)(3lnx+1)

So is it x^2 = 0 and

3lnx+1=0
11. (Original post by Indeterminate)
Note that, since

you have

which has TWO solutions
OP. according to my calculations Indeterminate is correct.
12. (Original post by Vorsah)
So once you get the2 values of X do u have to put them into the gradient function to see if they = 0

Eg for Q5 A I got x=0 or x=0.5 , so do I sub in these values into gradient function and see which one = 0 and then use the value that=0 and ignore the one that doesn't = 0?

When x=0, dy/dx=1

When x = 0.5 dy/dx=0 so this is the correct value
You differentiate the function to get dy/dy and equate that to zero. With 5a you get e^-2x(1-2x)=0 but e^-2x cannot equate to zero (ln0 is undefined) so you're left with 1-2x=0 so x=0.5. Find d^2y/dx^2 and sub x=0.5 to determine nature of point.
You differentiate the function to get dy/dy and equate that to zero. With 5a you get e^-2x(1-2x)=0 but e^-2x cannot equate to zero (ln0 is undefined) so you're left with 1-2x=0 so x=0.5. Find d^2y/dx^2 and sub x=0.5 to determine nature of point.
with 5A

I get x(e^-x) = x/(e^x)=0 >>> x=0 is this correct ?

and

2-x=0
14. (Original post by Vorsah)
with 5A

I get x(e^-x) = x/(e^x)=0 >>> x=0 is this correct ?

and

2-x=0
I'm not sure what you did to obtain that, what did you get when you did dy/dx? If you got x/(e^x) then that is incorrect. Use the product rule...

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