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
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
you ended up doing this, but your mistake was this: if then
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
while ...which is not a real number
draw a number line and place and 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!