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If a function is given whose domain and codomain are all real numbers such that

$$f(x) = x^3 + ax^2 + 3x + 100$$ then we have to find the value of $a$ for which the function is injective .

For that first I prove it as increasing or decreasing function . By doing this i got that the function is increasing when $a$ belongs to $[-3,3]$ and the function cannot be decreasing for any value of $a$.

But then I thought of second method by doing $f(x) =f(y)$. Solving this I got a equation of hyperbola i.e

$$x^2 +xy +y^2 + ax + ay +3 =0$$

But now how to proceed ?

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  • $\begingroup$ All polynomials of odd degree are automatically subjective by, for example, the IVT. Next, any strictly increasing continuous function must be injective, so just show that the function is strictly increasing $\endgroup$ – ASKASK Apr 11 '16 at 13:25
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$f$ is injective $\iff$ $\forall x,y \in D, f(x)=f(y) \Rightarrow x=y$

Assume $f$ is injective.

$$f(x)-f(y)=0 \Rightarrow (x-y)[x^2+xy+y^2+ax+ay+3]=0$$

As f is injective $x=y$ hence $3x^2+2ax+3$ is either equal to $0$ or not equal to $0$ for all $x$ in $\mathbb{R}$

The only possible case is $3x^2+2ax+3$ doesn't have any root in $\mathbb{R}$ This imples $$D<0 \Rightarrow 4(a^2-9)<0$$

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I would prefer your first method, but the second one can work too:

Actually your "hyperbola" is an ellipse when $|a|>3$ and nothing when $|a|<3$.

The function $$g(x,y) := \frac{f(x)-f(y)}{x-y} = x^2+xy+y^2+ax+ay+3$$ is an elliptic paraboloid opening upwards. You can find its bottom point by solving $\frac{\partial g}{\partial x} = \frac{\partial g}{\partial y}=0$.

If the value of $g$ at the bottom is less than zero, the locus of $g(x,y)=0$ must be a proper ellipse, which means there are nontrivial solutions to $f(x)=f(y)$.

On the other hand if the value of $g$ at the bottom is postive, then $g(x,y)=0$ can have no real solutions, and $f$ must be injective.

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$$f(x)=f(y) \Rightarrow x^3+ax^2+3x+100=y^3+ay^2+3y+100$$ $$(x^3-y^3)+a(x^2-y^2)+3(x-y)=0$$ $$(x-y)(x^2+xy+y^2+ax+ay+3)=0$$ Then $x-y=0$ and $x^2+xy+y^2+ax+ay+3 \not =0$ $$f(x)=x^2+x(y+a)+y^2+ay+3\not =0$$ $$D=(y+a)^2-4(y^2+ay+3)<0$$ $$-3y^2-2ay+a^2-12<0$$ $$3y^2+2ay-a^2+12>0$$ $$\frac D4=a^2+3(a^2-12)<0$$ $$4a^2-36<0$$ $$a^2<9 \Rightarrow a\in(-3;3)$$

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    $\begingroup$ You made a mistake when computing the derivative of $x^2$ and he already found the result using this method. $\endgroup$ – pmichel31415 Apr 11 '16 at 13:27
  • $\begingroup$ @Mandrathax please help me in second method. $\endgroup$ – user101522 Apr 11 '16 at 13:35
  • $\begingroup$ I edited my answer $\endgroup$ – Roman83 Apr 11 '16 at 13:42
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    $\begingroup$ @Roman83 I expected you would, I didn't downvote you ;) $\endgroup$ – pmichel31415 Apr 11 '16 at 13:46
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For your second method, you can write $$f(x)-f(y) = (x-y)\left[\frac12(x+y+\tfrac23a)^2+\frac12(x+\tfrac13a)^2+\frac12(y+\tfrac13a)^2 + \color{red}{\left(3-\tfrac13a^2\right)}\right]$$

Notice only if the last term in red is non-negative, the entire portion in square braces is never zero, so $f(x)-f(y) \implies x=y$. Thus our condition is $$a^2 \leqslant 9 \iff a \in [-3, 3]$$

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