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Let $\displaystyle f: R \rightarrow R $ be defined by $\displaystyle f(x) = x^3 + ax + 3 $ , where a is constant. Show that if $\displaystyle a \geq 0 $, then f is one-to-one.
I graphed this equation but I dont know what to do with it.... should I see its one to one.
To show something is one-to-one you need set x1=x2 and solve for x and try to get the same result on both sides....Quote:
Originally Posted by txsoutherngirl84
So you would:
x1^3+ax1+3=x2^3+ax2+3 <- now you solve for X
and try to end up with....
x1=x2 <- if you are successfully able do that then it IS 1 to 1.
Also, in your proof make sure to include "suppose that x1 and x2 are real numbers" and you need to satisfy the condition for ALL 'a' (so dont set a=to a number in your example).
btw,
Graphing 'one-to-one' and 'onto' problems help your understanding, but is a useless form of proof. If you are unsure if something is one-to-one or onto try to graph it. If it passes the 'vertical line' test it is one-to-one, if it passes the horizontal line-test is onto. So once you graph your line and run these tests then it can help you decide whether something is onetoone or onto. So then you know how to approach the problem and start your proof.
The part in red is incorrect. All functions pass the vertical line test, by definition. Passing the horizontal line test indicates one-to-one.Quote:
Originally Posted by matthayzon89
Edit: apparently the suggested method in the quoted text is also unworkable or not preferred, but it seemed plausible at first glance so I didn't say anything. Plato's method is very straightforward, so go with that.
Come on all of you: $\displaystyle f'(x)=3x^2+a\ge0$.
What does that tell you?
