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The function is defined as

$f : x \mapsto a + b \csc x$

$x \in \mathbb{R}$

$0 < x < 2\pi$

$x \neq \pi$

The stationary points of the graph $y = f(x)$ are

$(\frac{\pi}{2}, - 1)$ $(\frac{3\pi}{2}, - 5)$

There is a vertical asymptote at $x = \pi$.

What method would be used to find constants $a$ and $b$?

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Are you sure those stationary points are correct? –  Kirthi Raman May 23 '12 at 21:57
    
@tzxn3: Probable typo. –  André Nicolas May 23 '12 at 23:22

1 Answer 1

up vote 0 down vote accepted

I'll answer this assuming you have the stationary points

$(\frac{\pi}{2}, - 1)$ $(\frac{3\pi}{2}, - 5)$

As this seems most probable, either way, first draw a graph of $\csc x$ to get an idea of the way the function behaves if you don't know already, either way you will notice that the stationary points occur when $x = \frac{\pi(2n + 1)}{2}$ which agrees with our given points.

Now note that at these stationary points the values the function takes (the y coordinates) differ by 4, which tells us that our value of $b$. In order for the difference between stationary points to be twice that of the standard cosec function we need $b = 2$.

Now note that with $2 \csc x$ we still need to shift the function down in order for the stationary points to coincide with the ones given. But both points differ by the same amount giving us a. To take 2 to -1 and -2 to -5 we need to subtract 5, telling us that $a = -5$.

In summary to approach this sort of question look at how the function behaves on its own and try to see how you can transform the standard stationary points such that they all match, this will give you the constants relevant to the given function.

(If these points are not the ones in your original question, even better)

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