Question

A function f is defined by f: x-->e^x-1, where x>0
(1) State the range of f.
(2) Find an expression for f^-1
(3) State the domain of f^-1

Answer 1

Can anyone help me with this please?

Answer 2

e^x has a particular shape, where it is only valid when x>0 and y>0. so the problem is all set there

e^x - 1 means that the graph has been translated 1 up in the y-direction


this means the range of f is y>1
instead of y>0, because the graph has been moved up 1 in the y-direction

i am not sure what part 2 and 3 mean

Answer 3

I checked the marking scheme and it says the anser is > e–1 or > 0.37

Answer 4

question is for
\[e^x-1\] or
\[e^{x-1}\]?

Answer 5

will make a significant difference. part 2 is asking for the inverse function, but i cannot answer without knowing which one you mean

Answer 6

\[f(x)=e^x-1\] for x > 0 the range will be
\[(-1,\infty)\]

Answer 7

to find inverse set
\[x=e^y-1\] and solve for y via
\[x+1=e^y\]
\[y=\ln(x+1)\]
\[f^{-1}(x)=\ln(x+1)\]

Answer 8

No, the question is e^x-1 (e to the power of x-1) ;)