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$$\Rightarrow {x}^{2}+1={y}^{2}+1$$
$$\Rightarrow {x}^{2}={y}^{2}$$
$$\Rightarrow x=\pm y$$
This is not an unique solution. Here, “x” is not uniquely equal to “y”. We conclude that given function is not an injection. As a matter of fact, we can infer a check on our conclusion as,
$$\Rightarrow f\left(1\right)=f\left(-1\right)=2$$
Thus, we see that two pre-images relate to one image, which is contradictory to the requirement of an injection.
The fact that function value is different for different arguments has an important bearing on the nature of injection plot. Consider two plots shown in the figure. In the plot shown on the left, a straight line parallel to x-axis intersects the plot only once. In the second plot, a line parallel to x-axis intersects the plot at two points for $x={x}_{1}$ and $x={x}_{2}$ . The function represented by second plot is not an injection as two values of arguments map to a single value of function – not two different values as required for an injection function.
It means that intersection plot intersects a line parallel to x-axis only once. This is possible only if the function is either (i) continuously increasing or (ii) continuously decreasing. Note the use of word “continuously”. An increasing plot, for example, if drops, then we can always find a line parallel ot x-axis, which intersects it at two points.
Hence, an injection graph is either an increasing or decreasing type. We can associate these characteristics with differential calculus. We can say that :
Either
$$\frac{dy}{dx}>0\phantom{\rule{1em}{0ex}}\text{for all x}$$
or,
$$\frac{dy}{dx}<0\phantom{\rule{1em}{0ex}}\text{for all x}$$
As a matter of fact the derivative can be equal to zero for certain values of "x" - not for an interval of "x". Thus, we can write the condition of increasing function : iif function is continuous and
$$\frac{dy}{dx}\ge 0\phantom{\rule{1em}{0ex}}\text{for all x};\phantom{\rule{1em}{0ex}}\text{equality holding for certain values of x}$$
Similarly, we can write the condition of decreasing function : iif function is continuous and
$$\frac{dy}{dx}\le 0\phantom{\rule{1em}{0ex}}\text{for all x};\phantom{\rule{1em}{0ex}}\text{equality holding for certain values of x}$$
More than one pre-images of a function are related to same image.
The test of condition for many-one function is easy : if a function is not one-one, then it is many-one. Alternatively, we can check literally going by the definition – whether there exist such many-one relation. A map diagram showing the relation will reveal such instances of many-one relation.
Modulus function is one such many-one function. The function yields same value for positive and negative arguments of same magnitude.
$$f\left(x\right)=\left|x\right|$$
$$\Rightarrow f\left(-1\right)=|-1|=1$$
$$\Rightarrow f\left(1\right)=\left|1\right|=1$$
We should understand that a reverse function of the type “one to many” is barred from the very definition of function. The element of domain can be related to exactly one element in co-domain.
The definition of function puts the restriction on domain that every element in it is related. If we extend this restriction to co-domain also, then we get a function called “onto” or “surjection”.
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