# How to take the derivative of a function

To show that $f$ is 1-1, you could show that $$f(x)=f(y)\Longrightarrow x=y.$$ So, for example, for $f(x)=

Suppose $

I'll leave showing that $f(x)=<*is* 1-1 for you.

Alternatively, to show that $f$ is 1-1, you could show that $$x\ne y\Longrightarrow f(x)\ne f(y).$$

Or, to show that a differentiable $f$ is 1-1, you could show that its derivative $f'$ is either always positive or always negative.

You would discover that a function $g$ is not 1-1, if, when using the first method above, you find that the equation is satisfied for some $x\ne y$. For example, take $g(x)=1-x^2$. Then

$$ \eqalign< &g(x)=g(y)\cr \iff&<1-x^2>= <1-y^2> \cr \iff&-x^2= -y^2\cr \iff&x^2=y^2\cr> $$ The above equation has $x=1$, $y=-1$ as a solution. So, there is $x\ne y$ with $g(x)=g(y)$; thus $g(x)=1-x^2$ is not 1-1.

Of course, to show $g$ is not 1-1, you

need only find two distinct values of the input value $x$ that give $g$ the same output value.

Although you rightfully point out that the graphical method is unreliable; it is still instructive to consider the methods used and why they work:

Graphically, you can use either of the following:

Use the "Horizontal Line Test":

$f$ is 1-1 if and only if every horizontal line intersects the graph of $f$ in at most one point. Note that this is just the graphical interpretation of "if $x\ne y$ then $f(x)\ne f(y)$"; since the intersection points of a horizontal line with the graph of $f$ give $x$ values for which $f(x)$ has the same value (namely the $y$-intercept of the line).

Use the fact that a continuous $f$ is 1-1 if and only if $f$ is either strictly increasing or strictly decreasing. This, of course, is equivalent to the derivative being always positive or always negative in the case where $f$ is differentiable. (Note this method applies to only the green function below.)

Source: math.stackexchange.comCategory: Bank

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