Hi

Some questions I can't do. This one is Q6. The textbook answers are

x < - 2^0.5,x > 2^0.5

BTW I MEAN +/- ROOT 2 FOR THE LAST BIT OF WORKING

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- Oct 23rd 2007, 01:38 PMApexGeneral differentiation questions
Hi

Some questions I can't do. This one is Q6. The textbook answers are

**x < - 2^0.5**,**x > 2^0.5**

BTW I MEAN +/- ROOT 2 FOR THE LAST BIT OF WORKING

- Oct 23rd 2007, 01:47 PMJhevon
hint. a fraction is negative (that is less than zero) if we divide a negative number by a positive number, and vice versa (obviously, division by 0 is a no-no).

the denominator of f'(x) is a perfect square, thus it is always greater than or equal to zero. so to find where f'(x) < 0, we simply solve for where the numerator is < 0

so you want to solve $\displaystyle \left( x^2 + 2 \right) - 2x^2 < 0$

you ended up doing this, but your mistake was this: if $\displaystyle x^2 = 2$ then $\displaystyle x = \pm \sqrt{2} = \pm 2^{0.5}$ - Oct 23rd 2007, 02:07 PMApex
Hi

Is -2^0.5 equivalent to (-2)^0.5 :. wrong?

I don't see how to get to the final answer though. How do I know which way round the inequalities are, and whether the solutions are seperate from eachother?

Question 7 is below.

The textbook answer is x<4

Thanks v much

Apex - Oct 23rd 2007, 02:13 PMJhevon
yes, that is wrong.

$\displaystyle -2^{0.5} = - \sqrt{2}$ while $\displaystyle (-2)^{0.5} = \sqrt{-2}$ ...which is not a real number

Quote:

I don't see how to get to the final answer though. How do I know which way round the inequalities are, and whether the solutions are seperate from eachother?

- Oct 23rd 2007, 02:15 PMJhevon
- Oct 23rd 2007, 02:27 PMApex
Oh yeh thanks. I did it wrong way round. As for the numberline idea, I think there is a graphical method of doing this (how do my graphs help)? I need the graphical way if it's possible.

I've also tried Q8, which I get incorrect answers for.

Textbook answers are 0.25, 1.57, 2.89, 4.71.

Thanks so much! - Oct 23rd 2007, 02:39 PMApex
Hello,

This is question 9. Textbook doesn't say anything about it (it usually leaves "prove" answers out).

I'll post more questions tomorrow, unless I can somehow manage to complete them all correctly :P

Thanks :) - Oct 23rd 2007, 03:59 PMJhevon
the graphical way would be to graph $\displaystyle y = 2 - x^2$ and see the regions where y is less than zero (where the graph is under the x-axis, this is a lot more work than using the number line method however, usually, here, it's about the same, since 2 - x^2 is easy to graph)

- Oct 23rd 2007, 04:02 PMJhevon
in the line where you divided through by cos(x), don't. you may be dividing by zero (in fact, having cos(x) = 0 is one way to get the desired result).

$\displaystyle -2 \sin 2x + \cos x = 0$

$\displaystyle \Rightarrow - 4 \sin x \cos x + \cos x = 0$

$\displaystyle \Rightarrow \cos x ( -4 \sin x + 1 ) = 0$

$\displaystyle \Rightarrow \cos x = 0 \mbox { or } -4 \sin x + 1 = 0$

now continue - Oct 23rd 2007, 04:06 PMJhevon
you differentiated incorrectly

$\displaystyle y = x \sqrt{ \sin x }$

$\displaystyle \Rightarrow \frac {dy}{dx} = \sqrt{ \sin x } + \frac {x \cos x}{2 \sqrt{ \sin x}}$

(you forgot to use the chain rule when differentiating the $\displaystyle \sqrt{ \sin x}$)

now continue - Oct 24th 2007, 02:50 PMApex
Hi

I carried with Q9 but got stuck. See Q9 below.

Thanks alot

Apex - Oct 24th 2007, 02:53 PMApex
This one requires more of an explanation

- Oct 24th 2007, 02:53 PMApex
Stuck on solving this....

- Oct 24th 2007, 02:54 PMApex
I don't understand what this means.

- Nov 3rd 2007, 09:48 AMApex
bump