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November 25, 2014

November 25, 2014

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

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

Just looking at this gives me a headache. HELP!

- pre calculus -
**Damon**, Tuesday, December 20, 2011 at 8:31pmx = 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 8:41pmyour 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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