Okay I know the function $f(x)= x^3 +2x+3$ is one to one but how do I show that using the assumption that $f(a)=f(b)$ where $a=b$?
So if I do the process I have
$ a^3+2a+3 = b^3+2b+3$ my 3's cancel and i get
$a^3+2a= b^3+2b$
I know I can factor out an $a$ and $b$
$a(a^2+2)=b(b^2+2)$
but what do I do from here to show $a=b$?
$\endgroup$2 Answers
$\begingroup$Notice that
$$a^3+2a= b^3+2b\iff a^3-b^3+2(a-b)=0\iff (a-b)(a^2+ab+b^2+2)=0...(1)$$
Since $$a^2+ab+b^2+2=\left(a+\frac{b}{2}\right)^2+\frac{b^2}{4}+2\ge 2$$
it follows that the only real solution of ($1$) is $a=b$. Then $f$ is one-to-one.
Another approach:
Since $f'(x)=3x^2+2>0$ for any $x\in\mathbb{R}$ it follows $f$ is increasing, then monotone, which implies it is one-to-one.
$\endgroup$ $\begingroup$Let $f(a) = x^3 + 2x + 3$. IF $f(a) = f(b)$, you want to show that $a = b $. You can show this by contradiction. In fact, if $a \neq b$. Then, either $a > b$ or $a < b$. Consider first case: If $a > b $. Then $2a > 2b $ and $a^3 > b^3$ since cubic function is an increasing function. Hence,
$$ f(a) = a^3 + 2a + 3 > b^3 + 2b + 3 = f(b) \implies f(a) > f(b) $$
Contradiction since we assumed $f(a) = f(b)$
Similarly, if $b > a $ we obtain $f(b) > f(a) $, which is again a contradiction. IT follows that if $f(a) = f(b) $, then $a=b$.
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