General differentiation questions

**Apex** · October 23rd 2007, 01:38 PM

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

**Jhevon** · October 23rd 2007, 01:47 PM

Originally Posted by Apex

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

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 $\left( x^2 + 2 \right) - 2x^2 < 0$

you ended up doing this, but your mistake was this: if $x^2 = 2$ then $x = \pm \sqrt{2} = \pm 2^{0.5}$

**Apex** · October 23rd 2007, 02:07 PM

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

**Jhevon** · October 23rd 2007, 02:13 PM

Originally Posted by Apex

Hi

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

yes, that is wrong.

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

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?

draw a number line and place $\sqrt{2}$ and $- \sqrt{2}$ in their relative positions, then check numbers in the three regions in the formula for f'(x). so, for instance, you could try -2, 0 and 2. that way you can figure out what intervals work

**Jhevon** · October 23rd 2007, 02:15 PM

Originally Posted by Apex

Question 7 is below.
The textbook answer is x<4

Thanks v much
Apex

$8^{2/3} = 4$

**Apex** · October 23rd 2007, 02:27 PM

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!

**Apex** · October 23rd 2007, 02:39 PM

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

**Jhevon** · October 23rd 2007, 03:59 PM

Originally Posted by Apex

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.

the graphical way would be to graph $y = 2 - x^2$ 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)

**Jhevon** · October 23rd 2007, 04:02 PM

Originally Posted by Apex

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!

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).

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

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

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

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

now continue

**Jhevon** · October 23rd 2007, 04:06 PM

Originally Posted by Apex

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

you differentiated incorrectly

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

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

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

now continue

**Apex** · October 24th 2007, 02:50 PM

Hi

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

Thanks alot
Apex

**Apex** · October 24th 2007, 02:53 PM

This one requires more of an explanation

**Apex** · October 24th 2007, 02:53 PM

Stuck on solving this....

**Apex** · October 24th 2007, 02:54 PM

I don't understand what this means.

**Apex** · November 3rd 2007, 09:48 AM

bump

Thread: General differentiation questions

LinkBack

Thread Tools

Display

General differentiation questions

Similar Math Help Forum Discussions

General Combination Questions

General TI85 Questions

PDE - a few general questions

Some more general questions..

A few general questions and inquiries

Search Tags