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 November 30th, 2016, 08:50 AM #1 Member     Joined: Nov 2015 From: Alabama Posts: 84 Thanks: 9 Math Focus: Trigonometry, Calculus, Physics 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
 November 30th, 2016, 09:35 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,518 Thanks: 1240 $f(x) = x^3 + 2x + 1$ $f'(x) = 3x^2 + 2 > 0 \text{ for all } x \in \mathbb{R} \implies f(x) \text{ is strictly increasing}$ ... how would this help to show $f(x)$ is 1-1? Look at the definition of an injective function. Thanks from topsquark and SenatorArmstrong
November 30th, 2016, 09:47 AM   #3
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 Originally Posted by skeeter $f(x) = x^3 + 2x + 1$ $f'(x) = 3x^2 + 2 > 0 \text{ for all } x \in \mathbb{R} \implies f(x) \text{ is strictly increasing}$ ... 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.

November 30th, 2016, 09:55 AM   #4
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Quote:
 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?

November 30th, 2016, 11:31 AM   #5
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 Originally Posted by skeeter 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 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.

 December 13th, 2016, 09:55 PM #6 Senior Member   Joined: Sep 2016 From: USA Posts: 114 Thanks: 44 Math Focus: Dynamical systems, analytic function theory, numerics 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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