Question

Consider the graph of the function f(x)=e^x+1. The inverse of function f is ____ function. The inverse of function f has a domain of _____ and a range of ___.

an exponential
a linear
a quadratic
a logarithmic

all real numbers
x>0
x>1

y>1
all real numbers
y>0

Answer 1

Answer: Exponential, All real numbers, y>1

Explanation:

Answer 2

Here , the given function to us is
`{\bf{f(x)=e^(x)+1}}` which is clearly an exponential function , so it's inverse will be a logarithmic function . So , now let's solve for it's inverse

We have ;

`{:\implies \quad \sf f(x)=e^(x)+1}`

Now , let y = f(x) . So :

`{:\implies \quad \sf y=e^(x)+1}`

Now , solve for x

`{:\implies \quad \sf e^(x)=y-1}`

Take natural log on both sides :

`{:\implies \quad \sf ln(e^(x))=ln(y-1)}`

`{:\implies \quad \sf x\: ln(e)=ln(y-1)\quad \qquad \{\because ln(a^(b))=b\: ln(a)\}}`

`{:\implies \quad \sf x=ln(y-1)\quad \qquad \{\because ln(e)=1\}}`

Now , replace x by
`{\bf{f^(-1)(x)}}` and y by x

`{:\implies \quad \bf \therefore \quad \underline{\underline{f^(-1)(x)=ln(x-1)}}}`

Now , we got the inverse of f(x) which is a logarithmic function with domain
`{\bf{(-\infty , +\infty)}}` and Range
`{\bf{(1,\infty)}}`.So , the domain is the set of all real numbers and range being y > 1

Hence , The required answers are a logarithmic, all real numbers and y > 1