You are Here: Home >< Maths

# Core 3 Question - Parts (a) and (b) Watch

1. I need a step-by-step guide on answering Question 1(a) and (b) as follows.
Help would be much appreciated.

I attempted (a) and I know I've screwed up somewhere here.
I used the Product Rule for question (a).

y = e^-4x (x^2 + 2x - 2)

u = e^-4x
v = (x^2 + 2x - 2)

du/dx = -4e^-4x
Using the Chain Rule I got dv/dx = 2x + 2

Applying the Product Rule [dy/dx = u(dv/dx) + v(du/dx)]
I went onto get something different to the answer they wanted me to get, even through factorising. I've messed something up.

I would also like help on doing the second one. It's a five mark question and I don't know where to start as I can't recall C1 that well.

2. Well show what you got using the Product Rule (you didn't need chain rule btw).
3. You've got everything correct that you've written there, so your mistake will be after you've used the product rule. For the second part, the stationary points occur when dy/dx=0, so let dy/dx=0 and solve the equation to get the x values of the coordinates, then substitute it into the original equation to get the y values.
4. Yes I thought it was after the Product Rule too, writing up what I got is kind of long.

Could someone just complete the remainder of question (a) for me?
That way I'll know where I did actually mess it up
5. Oh and thanks a bunch for the replies!

6. Is what you have obtained using the product rule. If you take out a factor of then simplify the term in the bracket and take out another factor, then you should be able to obtain the answer in the required form.

7. For part B, just equal to 0
8. Thanks again.

I'm however still a bit stuck on the stationary points part though, I know I have to equal my dy/dx to 0 but what do I do after that to determine the co-ordinates?
9. (Original post by hass147)
Thanks again.

I'm however still a bit stuck on the stationary points part though, I know I have to equal my dy/dx to 0 but what do I do after that to determine the co-ordinates?
Solve the equation dy/dx=0 to find the x values of the coordinates, then substitute the x values into the original equation to find the y values of the coordinates.
10. (Original post by Phil_Waite)
Solve the equation dy/dx=0 to find the x values of the coordinates, then substitute the x values into the original equation to find the y values of the coordinates.
2e^-4x (5 - 3x - 2x^2) = 0

Divided both sides by 2e^-4x

(5 - 3x - 2x^2) = 0

-2x^2 - 3x + 5 = 0

-(-3) +/- root^[ (-3)^2 -4(-2)(5)]
_______________________
2(-2)

x = -5/2
x = 1

Is that correct so far?
Plugging them back into the equation confused me because I don't know if I'm supposed to substitute them into e^-4x's "x" or not and despite that, my answers come out as really small decimals or very high numbers. So I think something's amiss.
11. (Original post by hass147)
2e^-4x (5 - 3x - 2x^2) = 0

Divided both sides by 2e^-4x

(5 - 3x - 2x^2) = 0

-2x^2 - 3x + 5 = 0

-(-3) +/- root^[ (-3)^2 -4(-2)(5)]
_______________________
2(-2)

x = -5/2
x = 1

Is that correct so far?
Plugging them back into the equation confused me because I don't know if I'm supposed to substitute them into e^-4x's "x" or not and despite that, my answers come out as really small decimals or very high numbers. So I think something's amiss.
Yes that's correct, but you didn't need to use the quadratic formula because it easily factorises. Substitute them into all of the Xs in the original equation, for example:

Which simplifies to or and you should leave it in this form.
12. I was just doing this last night but I guess your question has been answered now. I went a different way about it personally but the result was the same.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 12, 2010
Today on TSR

### Degrees to get rich!

... and the ones that won't

### Women equal with Men?

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.