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# C3 Jan 07 Question watch

1. I'm stuck on edexcel Jan 07 question 2b and c...

I don't even understand what the question is asking me to do.

Click for the Jan 07 paper

I can't write it out because I'm not that great at using latex.

Thank you,
2. 2b. What's the minimum value of the function?

2c. What's the minimum value of the denominator of f(x)? Then use the fact that two positive numbers always divide to give a positive number.

Spoiler:
Show
If the minimum value is greater than 0 then whatever x is, x^2 + x + 1 will always be greater than 0
3. Complete the square for the numerator, that will tell you the range of the numerator (which will always be greater than 0.75, hence always greater than 0). Then C is just showing that since the numerator and denominator will always be positive, f(x) must be positive too...
4. b) put x^2+x+1/(x+2)^2 = 0
so x^2+x+1 = 0

as it is (x+2)^2, which cannot be negative because it is squared, therefore x^2+x+1 > 0

c) is asking to show x^2+x+1/(x+2)^2 is positive - basically because the denominator is square, and the numerator has a square (so they are of the same order) and as they are both squared, it has to be positive
5. So can I differentiate for 2b and subs the value of x into the equation?

Edit - Thanks for your replies. Rep coming your way in the next week or so.
6. no, just use x^2+x+1/(x+2)^2 = 0; x^2+x+1= 0 times (x+2)^2 therefore, x^2+x+1=0
7. 2b complete the square to show, 3 over 4 is your minimum value. therefore curve > 0

2c the denominator is a square, and we proved the numerator is positive, therefore f(x) > 0

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