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)
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!
2 answers
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) )
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) )
1 answer