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C3 help

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?

image.jpg224.5KB

Reply 1

Farhan.Hanif93

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

Reply 2

Indeterminate

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

e^{-x} \neq 0

you have

x(2-x)=0

which has TWO solutions :smile:

(edited 10 years ago)

Reply 3

Username_valid

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

Reply 4

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?

Reply 5

Username_valid

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

Reply 6

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

Reply 7

Vorsah

Original post by Username_valid

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

(edited 10 years ago)

Reply 8

The Shanus

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.

Reply 9

Vorsah

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

Reply 10

Pallas Athena

Original post by Indeterminate

Note that, since

e^{-x} \neq 0

you have

x(2-x)=0

which has TWO solutions :smile:

OP. according to my calculations Indeterminate is correct.

Reply 11

Username_valid

Original post by Vorsah

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.