# Graph drawing

1. Feb 28, 2013

### phospho

how can I draw y = arcsec(e^x) where x > 0 and 0 < y < pi/2

I've found the derivative which I got to be $\frac{1}{\sqrt{e^{2x} -1 }}$ however, the graph doesn't seem to have any minimum or maximum points.

I don't really know how to test this about, as I don't know what arcsec(e^x) is defined as, so I'm not sure how I can sketch it without being able to test some values of x..
thanks

2. Feb 28, 2013

### tiny-tim

hi phospho!
so you know it has no bumps!

ok, then all you need is to plot a few specimen values, and join them smoothly

(it might help calculations is you rewrite it as secy = ex, or x = … ?)

3. Feb 28, 2013

### HallsofIvy

I'm a bit confused as to what you are saying. arcsec is, of course, the inverse function to secant which is itself the reciprocal of cosine. If y= arcsec(x) then x= sec(y)= 1/cos(y). So cos(y)= 1/x and y= arccos(1/x). arcsec(e^x), then, is arccos(e^{-x}) and any calculator should have those functions.
For example, if x= 0, arcsec(e^x) is arccos(e^0)= arccos(1)= 0. If x= 1, arcsec(e^x)= arccos(e^{-1})= 0.9331, etc.

4. Feb 28, 2013

### pasmith

What you have there is $\frac1{\cos(y)} = e^x$, which can be re-arranged to give $\cos(y) = e^{-x}$ so that $x = -\ln(\cos(y))$ which might be easier to sketch.

5. Feb 28, 2013

### phospho

got it, thank you