# How would I show this bijection and also calculate its inverse of the function?

I want to show that $f(x)$ is bijective and calculate it's inverse.

Let $$f : \mathbf{R} \to \mathbf{R}$$ be defined by $f (x) = \frac{3x}{5} + 7$

I understand that a bijection must be injective and surjective but I don't understand how to show it for a function.

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A good place to start is with the definitions: Do you know what it means for a function $f$ to be injective? How about surjective? – user61527 Dec 3 '13 at 6:56
Yes I understand that an injection is for every element in the domain is mapped to one element is the co-domain. A surjection is that every element of the co-domain is mapped to at least one element of the domain. – Jack Dec 3 '13 at 6:59
Not really. The definition of a function is that one element can go to only one element of the codomain. An injective function is a function for which $f(x_1) = f(x_2) \implies x_1 = x_2$; that is, an element of the codomain can only come from one element of the domain. – user61527 Dec 3 '13 at 7:03
To restate @user61527’s explanation of injectivity, you need to verify that different $x$’s have different function values. – Lubin Feb 24 at 3:06

In a simple case like this, it may be that a geometric explanation is best. The graph $y=7 + 3x/5$ is a straight line, and since it intersects any horizontal line at most once, the function $f$ is injective. Since it intersects each horizontal line at least once, the function is surjective.

Example of an injective function that’s not surjective: $g(x)=e^x$. Example of a function that’s surjective without being injective: $h(x)=x^3-3x$. Example of a function that’s neither: $\sin x$.

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To show that $f(x)=\frac{3x}{5}+7$ is injective by the definition, note that $$\frac{3a}{5}+7=\frac{3b}{5}+7 \Rightarrow \frac{3a}{5}=\frac{3b}{5}\Rightarrow 3a=3b\Rightarrow a=b.$$ Therefore $f(x)$ is injective.

To find the inverse, trade $x$ and $y$ values in $y=\frac{3x}{5}+7$, and then solve for $y$: $$x=\frac{3y}{5}+7 \Rightarrow 5x=3y+35 \Rightarrow 3y=5x-35 \Rightarrow y=\frac{5x-35}{3}.$$

We can formally say that the inverse function of $f(x)$ is $$f^{-1}(x)=\frac{5x-35}{3}.$$

Regarding surjection: To show that the function is surjective, we want to show that for every real number $y$, there is a real number $x$ so that $f(x)=y$. A typical member of the domain $x$ is $x=\frac{5y-35}{3}$ (note we have not "traded" the $x,y$ values here from the original equation, but the process is the same). Since $y$ is a real number, certainly $\frac{5y-35}{3}$ is a real number. Now see that when we evaluate $f(x)$ at $x=\frac{5y-35}{3}$, we get

$$f\left(\frac{5y-35}{3}\right)=\frac{3\left(\frac{5y-35}{3}\right)}{5}+7=y.$$ This is also a real number. This means that for any real number $y$ we can always find an $x$ to "cover" that number.

Very specifically: You choose any real number $y$. I declare that $x=\frac{5y-35}{3}$ is the real value $x$ that covers that number so that $f(x)=y$, and because my value for $x$ is always defined for all real numbers $y$, we are safe. The $y$ values, or the "range", is "all used up", for lack of a more common phrase.

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So if I want to show that it is surjective I have to use the definition of y = f(x)? I am still confused as to how I would actually show the surjection. Also thanks for the help so far the injection actually makes sense to me along with the inverse. – Jack Dec 3 '13 at 7:16
I added a substantial edit to help. Definitely read (and try to truly understand) the definitions, they will help more. Let me know if this helps, or if anything is less than perfectly clear. – J. W. Perry Dec 3 '13 at 8:05
You implicitly used the property of (right) inverses that $f(f^{-1}(x)) = x$ in order to choose the value of $x$ that maps to $y$. Explaining that would make it clear why you chose $x = \frac{5y - 35}{3}$ when you were proving that $f$ is surjective. – N. F. Taussig Jan 22 at 14:51