Just checking on this one.

[Lizzie]

OK, I am given 3 graphed lines.
The blue line is y=cosx, the red line is y=sinx, and the green line is y=-sinx. Now I am supposed to do this...

"The following figure shows the graphs of f, f', and F (where F'=f). Identify each of these by color."

My answer was that f was the blue graph (y=cosx), f' was the green graph (y=-sinx) and F was the red graph (y=sinx).

I am pretty sure that I am right.

Just in case I am wrong and so that I don't frustrate anyone, here is exactly how I did this.

I knew that F' was the same as f, so I tried to see which one was the derivative of which. I found what each line's equation was and figured out that the derivative of y=sinx is y'=cosx. So, I figured that y=sinx was F, and that y=cosx was f. Then I found that the other line was y=-sinx. I took f, which was y=cosx and found the derivative, which is y=-sinx, which is my last line.

I hope that wasn't too jumbled and confusing. Anyway, I think that i did this one right.

 

[soroban]

Hello, Lizzie!

You're dead-on!

I am given 3 graphed lines.
The blue line is \(\displaystyle y\,=\,\cos x\), the red line is \(\displaystyle y\,=\,\sin x\), and the green line is \(\displaystyle y\,=\,-\sin x\)

The following figure shows the graphs of \(\displaystyle f,\;f',\;\) and \(\displaystyle F\) (where \(\displaystyle F'\,=\,f\)).
Identify each of these by color.

My answer was:
\(\displaystyle f\,(y\,=\,\cos x)\) was the blue graph,
\(\displaystyle f'\,(y\,=\,-\sin x)\) was the green graph,
and \(\displaystyle F\,(y\,=\,\sin x)\) was the red graph.

I am pretty sure that I am right. . You are . . . and your reasoning is excellent!!

Nice work!

 

[Lizzie]

YAY! Thanks!