I have no idea where to begin. I'm not used to these kinds of problems.

Could someone solve it/show work. Please and thank you.

http://img296.imageshack.us/img296/2...problemmn8.png

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- Jun 23rd 2008, 12:45 PMMathNubecosine
I have no idea where to begin. I'm not used to these kinds of problems.

Could someone solve it/show work. Please and thank you.

http://img296.imageshack.us/img296/2...problemmn8.png - Jun 23rd 2008, 12:52 PMicemanfan
So you want the value of k such that $\displaystyle \cos x = x^2 + k$ at the value of $\displaystyle x = \frac{\pi}{2}$. This is the equation: $\displaystyle \cos {\frac{\pi}{2}} = \left({\frac{\pi}{2}}\right)^2 + k$. This reduces to $\displaystyle 0 = \frac{\pi^2}{4} + k$. Hence, $\displaystyle k = -\frac{\pi^2}{4}$.

- Jun 23rd 2008, 02:13 PMMathNube
- Jun 23rd 2008, 02:16 PMtopsquark
For the function to be continuous the values of the two branches of the function must be equal where they meet. Otherwise the function will suffer a "jump" (ie. discontinuity) at x = pi/2. Thus we require

$\displaystyle cos(x) = x^2 + k$ at $\displaystyle x = \frac{\pi}{2}$

And the solution goes on from there as icemanfan wrote.

-Dan - Jun 23rd 2008, 02:16 PMIsomorphism
- Jun 23rd 2008, 02:34 PMMathNube
ok, why does x=pi/2

- Jun 23rd 2008, 02:35 PMtopsquark
- Jun 23rd 2008, 04:32 PMMathNube
Are we positive that this is the right answer? According to the person who gave me the problem it is incorrect.

- Jun 23rd 2008, 05:40 PMtopsquark
- Jun 23rd 2008, 05:52 PMMathNube
I was not given the solution, just told that it was incorrect.

- Jun 23rd 2008, 06:05 PMtopsquark
I find it hard to argue with the graph. The graphs of the two functions on their respective intervals is clearly continuous with the choice k = -pi^2 / 4. I don't see any way that this solution can be wrong.

Show this graph to whoever gave you the problem and ask them what their solution looks like. If we are wrong I'd like to see just what the answer is.

-Dan