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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

Consider The Graph Of The Function Fxex1 The Inverse Of Function F Is Function The Inverse Of Function F Has A Domain Of And A Range Of An Exponential A Linear class=

Sagot :

Medunno13

Answer: Exponential, All real numbers, y>1

Step-by-step explanation:

Here , the given function to us is [tex]{\bf{f(x)=e^{x}+1}}[/tex] 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 ;

[tex]{:\implies \quad \sf f(x)=e^{x}+1}[/tex]

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

[tex]{:\implies \quad \sf y=e^{x}+1}[/tex]

Now , solve for x

[tex]{:\implies \quad \sf e^{x}=y-1}[/tex]

Take natural log on both sides :

[tex]{:\implies \quad \sf ln(e^{x})=ln(y-1)}[/tex]

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

[tex]{:\implies \quad \sf x=ln(y-1)\quad \qquad \{\because ln(e)=1\}}[/tex]

Now , replace x by [tex]{\bf{f^{-1}(x)}}[/tex] and y by x

[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{f^{-1}(x)=ln(x-1)}}}[/tex]

Now , we got the inverse of f(x) which is a logarithmic function with domain [tex]{\bf{(-\infty , +\infty)}}[/tex] and Range [tex]{\bf{(1,\infty)}}[/tex].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

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