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Prove that the following functions defined in $R\to R$ are neither injective nor surjective.

$(i)(x^2+x+5)(x^2+x-3)\hspace{2cm}(ii)\frac{x^2+4x+30}{x^2-8x+18}$

Since the injective test says if $f(x_1)=f(x_2)\Rightarrow x_1=x_2$,then $f(x)$ is injective otherwise not.And the Surjective test says that for every $y\in R$,there is a $x\in R$.

When i apply above injective test,the simplification goes messy and does not come $x_1=x_2$ and surjective test is also not working for these functions.

When i graphed these functions on desmos.com graphing calculator,i can see that it is not injective because horizontal test fails.But how to check surjective by looking at the graph?

Are these two methods available or other methods are there for checking surjectivity and injectivity of these type of complicated functions.

Please help me.

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Of course the test does not simplify to $x_1 = x_2$, because you are not supposed to prove that the function is injective, you should prove it is not injective. To do that, you can either find some $x_1\neq x_2$ such that $f(x_1)=f(x_2)$ or at least prove that such a pair exists.

I suggest you look at the limits of both functions as $x\to\infty$ and $x\to -\infty.$ Using these limits, you can show that the function is not injective, without actually finding a particular pair of points.

On the other hand, you can always simply pick some number, like $x_1=0$, and then prove that the equation $f(x)=x_1$ has two distinct solutions. Unless you were very unlucky in picking $x_1$, that should not be hard.


For surjectivity, you need to prove that for every $y$, there exists some $x$ that $f(x)=y$. In your case, you need to disprove surjectivity, so you must find some $y$ for which $f(x)$ is never equal to $y$. Graphically, you need to find a number $y_0$ such that the line $y=y_0$ never intersects with the graph of $f$.

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  • $\begingroup$ How does finding the limit at $\pm\infty$ prove the function injective.Is it always applicable.For my first function,limit at infinity is infinity. @5xum $\endgroup$ – diya Sep 16 '15 at 12:10
  • $\begingroup$ @diya Because the limit $\lim_{x\to\infty} f(x) = \infty$, that means that for every $M\in\mathbb R$, there exists such a $x_M>0\in\mathbb R$ that $f(x) > M$ if $f(x)>M$. Same for negative values and $x_M<0.$ $\endgroup$ – 5xum Sep 16 '15 at 12:12
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    $\begingroup$ @Svetoslav If both limits are $\infty$ and the function is continuous, then the function is not injective. Also, if both limits are $L$ and the function is continuous and constantly larger or smaller than $L$, then the function is not injective. $\endgroup$ – 5xum Sep 16 '15 at 12:16
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    $\begingroup$ @diya There are no rules for this kind of task. You think about it hard enough and try a couple of approaches, and then one sticks. Then you repeat the same thing on $50$ different exercises and you notice that you can now solve the task in one tenth of the time as before. That's when you know you practiced enough. $\endgroup$ – 5xum Sep 16 '15 at 12:29
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    $\begingroup$ @diya You are overcomplicating things. If $f$ is a continuous function and the limit of $f(x)$ as $x\to\pm \infty$ exists (and is equal in both cases), you have three options: (1) The limit is $\infty$, in which case $f$ is bounded below (2) The limit is $-\infty$, in which case $f$ is bounded above (3) The limit is finite, in which case $f$ is bounded from both sides. $\endgroup$ – 5xum Sep 16 '15 at 12:52

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