One to one or injective means that each x in the domain maps to a unique y, i.e. the function preserves distinction.
y=(x-1)/(3x+3) is one to one. We can solve for x:
3xy + 3y = x - 1
3y + 1 = x - 3xy = x(1-3y)
x = (3y+1)/(1-3y)
That proves it's one-to-one; for each y we get one x.
[tex]y=\sqrt{5x+9}[/tex]
That's also one to one,
[tex]y^2 = 5x+9[/tex]
[tex]x = \frac 1 5 (y^2 - 9)[/tex]
That's the proof.
[tex]f(x)=\dfrac{7}{4x^2}[/tex]
That's Not one-to-one
because f(-x)=f(x). We only need a single counterexample; f(1)=7/4, f(-1)=7/4
[tex]f(x) = \frac 1 2 x^3[/tex]
That's one-to-one, textbook case
[tex]f(x)=3x^4 + 7x^3[/tex]
That's not one-to-one because the fourth power dominates so this will be sort of parabola-ish, repeating the same big positive values for a negative and positive x.
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[tex]y = f(x) = \sqrt[3]{x-2} + 8[/tex]
[tex]y - 8 = \sqrt[3]{x-2}[/tex]
[tex](y-8)^3 = x-2[/tex]
[tex]x = (y - 8)^3 + 2[/tex]
[tex]f^{-1}(x) =(x - 8)^3 + 2[/tex]
First choice
