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Question: Which of the following are injections, or surjections or bijective? Justify your answer. \(f:\left( ...

Which of the following are injections, or surjections or bijective? Justify your answer. f:(0,)Rf:\left( 0,\infty \right)\to \text{R} defined by f(x)=logexf\left( x \right)={{\log }_{e}}x.

Explanation

Solution

We must first assume two variables x1 and x2{{x}_{1}}\text{ and }{{x}_{2}} is the domain of given function, such that f(x1)=f(x2)f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right). If we get x1=x2{{x}_{1}}={{x}_{2}}, then the function is injective. Again, assuming a variable y1{{y}_{1}} in codomain, such that f(x1)=y1f\left( {{x}_{1}} \right)={{y}_{1}}, and if we get that for every y1{{y}_{1}}, the corresponding x1{{x}_{1}} lies in the domain, then the function is surjective. We must also know that if the function is both injective and surjective, it is called bijective function.

Complete step-by-step solution:
Let us assume two values x1,x2(0,){{x}_{1}},{{x}_{2}}\in \left( 0,\infty \right) such that f(x1)=f(x2)f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right).
Using the definition of this given function, we can write
logex1=logex2{{\log }_{e}}{{x}_{1}}={{\log }_{e}}{{x}_{2}}
We can also write this as
logex1logex2=0{{\log }_{e}}{{x}_{1}}-{{\log }_{e}}{{x}_{2}}=0
Using the property of logarithms, we can write
logex1x2=0{{\log }_{e}}\dfrac{{{x}_{1}}}{{{x}_{2}}}=0
On taking exponential on both sides, we can write
elogex1x2=e0{{e}^{{{\log }_{e}}\dfrac{{{x}_{1}}}{{{x}_{2}}}}}={{e}^{0}}
Again, using the property of logarithms, we can write
x1x2=1\dfrac{{{x}_{1}}}{{{x}_{2}}}=1
And so, we get
x1=x2{{x}_{1}}={{x}_{2}}.
Hence, the given function is one-one. Or, we can say that the given function is injective.
Let us now assume a variable y1R{{y}_{1}}\in \text{R}, such that f(x1)=y1f\left( {{x}_{1}} \right)={{y}_{1}}.
Thus, we get
logex1=y1{{\log }_{e}}{{x}_{1}}={{y}_{1}}
Using the definition of logarithms, we can write
x1=ey1{{x}_{1}}={{e}^{{{y}_{1}}}}
We can see here that for every y1R{{y}_{1}}\in \text{R}, the corresponding x1(0,){{x}_{1}}\in \left( 0,\infty \right).
Thus, we can say that the function is onto or surjective.
We now know that the given function is both injective and surjective.
Hence, we can say that the function is bijective.

Note: We must carefully include all the boundary conditions while checking the function for injectivity and for surjectivity. We must also be very clear that the injective function is also known as one-one function, and the surjective function is also called an onto function.