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November 30th, 2016, 08:50 AM   #1
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Best way to show that a function is one-to-one

Hello.

I am trying to find out what would be the best way to show that a function is one-to-one without graphing and doing the horizontal line test. Would I just perform a concavity test for all values on a designated interval?

For example, say I have f(x) = x^3 + 2x + 1 .

How could I show that this function is or is not one-to-one?

Many thanks

Jacob
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November 30th, 2016, 09:35 AM   #2
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$f(x) = x^3 + 2x + 1$

$f'(x) = 3x^2 + 2 > 0 \text{ for all } x \in \mathbb{R} \implies f(x) \text{ is strictly increasing}$

... how would this help to show $f(x)$ is 1-1?

Look at the definition of an injective function.
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November 30th, 2016, 09:47 AM   #3
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Quote:
Originally Posted by skeeter View Post
$f(x) = x^3 + 2x + 1$

$f'(x) = 3x^2 + 2 > 0 \text{ for all } x \in \mathbb{R} \implies f(x) \text{ is strictly increasing}$

... how would this help to show $f(x)$ is 1-1?

Look at the definition of an injective function.
Got it. So taking the derivative and then set it to be greater than zero. Doing this we can see that the derivative is never less than zero. Which means it is positive and therefore always increasing. It will always pass the horizontal line test for all real numbers.

Appreciate it.
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November 30th, 2016, 09:55 AM   #4
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So taking the derivative and then set it to be greater than zero.
you're not "setting it greater than zero" ... you're making the observation that $f'(x) = 3x^2+2$ is always greater than zero.

btw, a function $f(x)$ would also be 1-1 if $f'(x) < 0$ for all $x$ in its domain ... true?
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November 30th, 2016, 11:31 AM   #5
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you're not "setting it greater than zero" ... you're making the observation that $f'(x) = 3x^2+2$ is always greater than zero.
Ahh good point.

Quote:
Originally Posted by skeeter View Post
btw, a function $f(x)$ would also be 1-1 if $f'(x) < 0$ for all $x$ in its domain ... true?
And funny that you mention that. I am in the math lab right now and that crossed my mind. I then came onto mmf and here you are pointing it out. But yes! That would make sense.
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December 13th, 2016, 09:55 PM   #6
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Suppose $f(a) = f(b)$ and WLOG assume $a \leq b$. Then $a^3 + 2a = b^3 + 2b$ which can be rewritten as $a^3 - b^3 + 2(a-b) = 0$. Factoring $(a-b)$ you get $(a-b)(a^2 + ab+b^2 + 2) = 0$ and since $b^2 \geq ab$ it follows that $a^2 + ab + b^2 + 2 > 0$ implying $a-b = 0$.
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