Are these invertible? Why or why not?

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a.) sech x on (0,infinity)

b.) cos (ln x) on (0, e^pi)

c.) e^(x^2) on (-1,2)

I am stuck and have no clue. I have only been able to get through basic questions like this, how do you complete these?

Thanks for any help.
 
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whats the definition of invertible? good place to start...

welcome to pf by the way ;)

it generally works here by you having an attempt & and people will help steer you through the problem - though you still do the work
 
The usual strategy is to sketch a graph of each function and try to figure out whether each horizontal line intersects the graph at most once. What do you say for those three examples?
 
Quick question - how do you plot a function like sechx, cos(lnx), etc with a scientific calculator?
Do you need to actually rearrange to see if its invertible or is there a other method? (aka just by looking?)

a) sechx is not 1-1 (can tell from plotting it)

b) Not sure

c) not 1-1 (from graph)
 
adelaide87 said:
Quick question - how do you plot a function like sechx, cos(lnx), etc with a scientific calculator?
Do you need to actually rearrange to see if its invertible or is there a other method? (aka just by looking?)

a) sechx is not 1-1 (can tell from plotting it)

b) Not sure

c) not 1-1 (from graph)

I'm not much on calculators so I can't answer the first question, but if you plotted sech(x) then, yes, it's not invertible, BUT you are only looking at the (0,infinity) part. Do you want to rethink that opinion? To prove a function is NOT invertible you only need to find two values of x, say x1 and x2 such that f(x1)=f(x2).
 
Indeed, with sech(x), you need to look only at the positive numbers. Be careful there.

Also, cos(ln(x)) is easy to do on a TI calculator. Just make sure you know your bounds.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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