Posted by **Michelle** on Tuesday, December 20, 2011 at 8:10pm.

Find the inverse of the function below. Graph the function below and the inverse. Determine the domain, range and asymptotes of the function below and the inverse function. Please show all your work.

f(x) = 2e^-x + 5

Just looking at this gives me a headache. HELP!

- pre calculus -
**Damon**, Tuesday, December 20, 2011 at 8:31pm
x = 2 e^-y + 5

2 e^-y = x-5

e^-y = (x-5)/2

ln e^-y= -y = ln [(x-5)/2]

y = - ln [(x-5)/2]

= - [ ln(x-5) - ln 2 ]

y = ln 2 - ln(x-5)

- pre calculus -
**Reiny**, Tuesday, December 20, 2011 at 8:41pm
your given function is

y = 2e^-x + 5

to form the inverse, interchange the x and y variables, so the inverse is

x = 2e^-y + 5

the inverse graph will be a reflection of the original graph in the line y = x

Pick a few ordered pairs of the original function, e.g.

(0,7), (1, 5.7) , (3, 5.1) , (-1, 10.4 ) , (-5, large) , (5, just a bit over 5)

sketch the first graph

for the inverse, switch the x and y of the ordered pairs, (same as refection in line y = x)

if you want to express your inverse as a function....

x-5 = 2e^-x

take ln of both sides

ln (x-5) = ln2 + ln e^-y

ln(x-5) - ln2 = -ylne , but lne = 1

y = ln2 - ln(x-5)

or

y = ln(2/(x-5) )

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