ya, i know, i though no one visited the mathematics section, because i was only able to join it by using the search tool. Also, i put a note for someone to delete my post in the other section.:smile:
btw, i posted back on you reply.:tongue2:
Thats what I got, but I couldnt figure out how to solve the tan 2x = 2x. How would I be able to solve it?
I tried to graph it, but the points that came out, did not work in the equation.:yuck:
So, do I go with x=0 only?
edit: my teacher said, I was gonna use decimals on this problem.
edit...
Well, I got this equation f(x)=\frac{2\sin{2x}}{x} [-\pi,\pi]
So I took the 1st derivative, f'(x)=\frac{2(2x\cos{2x} - \sin{2x})}{x^2}
Then I set that equal to 0, and got 0=2x\cos{2x} - \sin{2x}
But I do not see how to get the critical numbers, I also tried to do double angle, but that...
Well, I got this equation f(x)=\frac{2\sin{2x}}{x} [-\pi,\pi]
So I took the 1st derivative, f'(x)=\frac{2(2x\cos{2x} - \sin{2x})}{x^2}
Then I set that equal to 0, and got 0=2x\cos{2x} - \sin{2x}
But I do not see how to get the critical numbers, I also tried to do double angle, but that just...
a= (-6/a^2) - (1/2)??????
i dont know, seems like this question could take up a long, long time. this is supposedly an ap free response question. im thinking this question, is a little too long, to be on a timed test.
hmmm, so
b = 3a^3 - 2a^2 - 4a - 8
b = a^3 - a^2 - 4a + 4
0=-2a^3 +a^2 +12
yuck, im sure this wasnt suppose to happen. :(
edit: somehow, i think im making it harder than it seems, looks like a very simple problem. :\
This somehow seems easy, but I cannot get the jist of it.
f(x) = x^3 - x^2 - 4x + 4
The point (a,b) is on the graph of f and the line tangent to the graph at (a,b) passes through the point (0,-8) which is not on the graph of f. Find the values of a and b.
okay so, ive spent about an...