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 $\displaystyle \left( x^2 + 2 \right) - 2x^2 < 0$
you ended up doing this, but your mistake was this: if $\displaystyle x^2 = 2$ then $\displaystyle x = \pm \sqrt{2} = \pm 2^{0.5}$
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
yes, that is wrong.
$\displaystyle -2^{0.5} = - \sqrt{2}$ while $\displaystyle (-2)^{0.5} = \sqrt{-2}$ ...which is not a real number
draw a number line and place $\displaystyle \sqrt{2}$ and $\displaystyle - \sqrt{2}$ 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 workI 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?
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!
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)
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