Science:Math Exam Resources/Courses/MATH220/December 2009/Question 04 (a)

MATH220 December 2009

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Question 04 (a)
Prove that $f:[3,\infty )\to [5,\infty )$ defined by $\displaystyle {}f(x)=x^{2}-6x+14$ is a bijection.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Completing the square helps make the argument very routine.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The function defined by $\displaystyle f(x)=x^{2}-6x+14=x^{2}-6x+9-9+14=(x-3)^{2}+5$ (where above we completed the square) will map 3 to 5 and anything larger than 3 to something larger than 5. To show that this function is injective, we suppose that there exist real numbers $\displaystyle x,y\in [3,\infty )$ such that $\displaystyle (x-3)^{2}+5=(y-3)^{2}+5$ Simplifying shows that $\displaystyle x-3=\pm (y-3)$ Since the left hand side above is positive (as x is at least 3), so is the right hand side. As $\displaystyle y-3$ is positive, we can take the positive sign on the right. Thus $\displaystyle x-3=y-3$ and simplifying shows that x and y must be equal. This shows that our function is an injection. To show that it is a surjection, we need to show that for any value c inside $\displaystyle [5,\infty )$, there is some value $\displaystyle x\in [3,\infty )$ such that $\displaystyle f(x)=(x-3)^{2}+5=c$ If we isolate for x above, we see that $\displaystyle x=3\pm {\sqrt {c-5}}$ If we take the positive sign above, we see that this x values satisfies $\displaystyle f(x)=c$ completing the claim that f is a surjection. Thus, as our function is injective and surjective, it is a bijection and this completes the proof.

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