Inverse Equation and if Values exist

[johnq2k7]

Please help me with these following problems:

1.)Indicate whether each of the following functions is invertible in the given interval. Explain

a.) sech x on [0,infinity)

b.) cos (ln x) on (O, e^pie]

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


My work process for a: let y= sech x

how do i find x in terms of y for this eq. how do i isolate the x variable from sech x.

once I find that i can see if the positive values of x for the inverse function exist from 0 to infinity.

however, i need help finding the inverse function


My work process for b.) let y= cos (ln x)

I took the expotential of both sides to get

e^y= e^(cos*(lnx))

how do i isoloate e*(lnx) to equal x , since the lnx is a part of the cosine angle expression.. i need help here...

once i find the inverse equation, i can see if y values exists for the x values from 0 to e^pie... i need help finding the inverse equation first



My work process for c.) let = y= e^(x^2)

therefore taking the natural logarithm of both sides i get


ln y= ln (e^(x^2)
ln y= x^2

therefore x= sqrt (ln y)

therefore f^-1 (x)---> inverse function is f(x)= sqrt (ln x)

when subbing (-1,2] for x to determine if y values exist,

i find the inverse function doesn't exist under those limits since ln 0 is undefined

is this the correct approach to this problem.


Please help me with these problems!

 

[HallsofIvy]

A function is invertible in an interval if it is "one-to-one" there: that is, that two different values of x in the interval that give the same value of y. It should be fairly obvious, for example, that (-.5)²= .5² so $e^{(-.5)^2}= e^(0.5)^2$.

You say at one point

ln y= x^2
therefore x= sqrt(ln y)

which is not true: if ln y= x^2 then either x= sqrt(ln y) or x= -sqrt(ln y).

Your point about ln(0) not existing is a good one.

For the others try looking at graphs of the functions.