Question

Need help with equations! :)
See picture!

Answer 1

4=2^2
Right ?
So we have :
2^-x^2+x+2=2^2
We have equal bases : 2 = 2
So have an equation with powers :
-x^2+x+2=2
No find x .

Answer 2

Now not No .

Answer 3

Now find x .

Answer 4

So
\[-x ^{2}+x <0\]
and then
\[x = 0\] and \[x=1\] ?

Answer 5

And btw, do I use x=0 or x<0
it's been so long since I've done this type of math :P

Answer 6

And this one is even more troubling for me! :P
\[(-x^{2}-1)\sin(2x) \ge 0 , for, x \in \left[ 0,2 \pi \right]\]

Answer 7

once you get to
\[ -x ^{2}+x <0 \]
factor out an x
\[ x ( -x + 1) < 0 \]
or
\[ x(1-x) < 0 \]
the expression on the left is negative (i.e. less than zero) if one of the terms is negative and the other is positive (because minus * plus is minus)
we need either
x < 0 and 1-x > 0
or
x>0 and 1-x < 0

one of those sets should be possible.

Answer 8

For
\[ (-x^{2}-1)\sin(2x) \ge 0 , \text{ for } x \in \left[ 0,2 \pi \right] \]
we use the same idea: first factor out the minus sign
\[ -(x^2+1) \sin(2x) ≥ 0 \]

x^2 + 1 is always positive, and -(x^2+1) is always negative.
to get a positive number on the left we want sin(2x) ≤ 0 (negative)
we need all angles where sin(2x) is negative.

Answer 9

Wow, that was incredibly well explained! :D Thanks!^^

Answer 10

But I'm still stuck :P I don't even know where to go from there :/

Answer 11

sin 2x will have two periods in the interval 0 to 2pi
so there will be two intervals where sin 2x is negative. The first interval is
\[ \pi/2 ≤ x ≤ \pi \]

Answer 12

you type
plot sin(2x)
into the google search window to see a graph.
remember pi is about 3.14 and 2pi is about 6.3

Answer 13

So the other one is \[-\frac{ \pi }{ 2} \le x \le 0\] ?

Answer 14

they restricted x to the interval 0 to 2pi

But you can add 2pi to your answer to get the correct interval.
2pi - pi/2 to 2pi
or
3pi/2 to 2pi is the 2nd interval