Show that f : R R defined by f(x) = cx + d where c( 0) and d are real numbers, is one to one and onto. Find f^ 1(x) .

Solution

1. Proving $f$ is One-to-One (Injective)

A function $f:\mathbb{R}\to \mathbb{R}$ is one-to-one (injective) if for any $x_1,x_2\in \mathbb{R}$, $f(x_1)=f(x_2)$ implies $x_1=x_2$.

Let $x_1,x_2\in \mathbb{R}$ such that: $$f(x_1)=f(x_2)$$ Given $f(x)=cx+d$, we substitute: $$c{x}_1+d=c{x}_2+d$$ Subtract $d$ from both sides: $$c{x}_1=c{x}_2$$ Since $c\ne 0$, we can divide both sides by $c$: $$x_1=x_2$$ Thus, $f$ is one-to-one.

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2. Proving $f$ is Onto (Surjective)

A function $f:\mathbb{R}\to \mathbb{R}$ is onto (surjective) if for every $y\in \mathbb{R}$, there exists an $x\in \mathbb{R}$ such that $f(x)=y$.

Let $y\in \mathbb{R}$ be arbitrary. We need to find $x\in \mathbb{R}$ such that: $$f(x)=y$$ Substitute $f(x)=cx+d$: $$cx+d=y$$ Solve for $x$: $$cx=y-d$$ $$x=\frac{y-d}{c}$$ Since $c\ne 0$ and $y,d\in \mathbb{R}$, $x$ is well-defined and real. Therefore, for every $y\in \mathbb{R}$, there exists an $x\in \mathbb{R}$ such that $f(x)=y$. Thus, $f$ is onto.

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3. Finding the Inverse Function $f^{-1}(x)$

Since $f$ is both one-to-one and onto, it is bijective, and its inverse $f^{-1}$ exists.

From the surjectivity proof, we derived: $$x=\frac{y-d}{c}$$ To express $f^{-1}(x)$, replace $y$ with $x$: $$f^{-1}(x)=\frac{x-d}{c}$$

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Final Answer

The function $f(x)=cx+d$ ($c\ne 0$) is one-to-one and onto, and its inverse is: $$f^{-1}(x)=\boxed{\frac{x-d}{c}}$$