Given two sets X and Y, a function from *X* to *Y* is a rule, or law, that associates to every element *x* ∈ *X *(the independent variable) an element *y* ∈ *Y* (the dependent variable). It is usually symbolized as

* *

in which *x* is called argument (input) of the function *f* and *y* is the image (output) of *x* under *f*.

A single output is associated to each input, as different input can generate the same output.

The set *X* is called domain of the function *f* (dom f), while Y is called codomain (cod f). The range (or image) of *X*, is the set of all images of elements of *X* (rng ƒ). Obviously

The notation

indicates that ƒ is a function with domain *X *and codomain *Y*.

Given ƒ:X → Y, the **graph** *G( f )* is the set of the ordered pairs

In particular, if *x* and *y * are real numbers, *G(f )* can be represented on a Cartesian plane to form a curve. A glance at the graphical representation of a function allows us to visualize the behaviour and characteristics of a function.

Given *ƒ:X → Y*, the **preimage** (or **inverse image**, or **counter image**) of a subset *B* of the codomain *Y* under ƒ is the subset* f ^{-1}*(B) of the elements of

The Vertical line test is used to determine whether a curve is the graph of a function when the function’s domain and codomain correspond to the *x* and *y* axes of the Cartesian coordinate system.

Essentially, the test amounts to answering this question:

is it possible to draw a vertical line that intersects the curve in two or more places? If so, then the curve is not the graph of a function. If it is not possible, then the curve is the graph of a function.

Why does this test work?

A curve would fail to be the the graph of a function if for any input x, there existed more than one *y*-value corresponding to it.

Take, for example, the equation

Note that the points (0, 2) and (0, -2) both satisfy the equation. So we have a situation in which one *x*-value (namely, when *x* = 0) corresponds to two different y-values (namely, 2 and -2). The points (0, -2) and (0, 2) lie on the same vertical line with equation *x* = 2 on the Cartesian coordinate system.

So if a vertical line hits a curve in more than one place, it is the same as having the same *x*-value paired up with two different *y*-values, and the graph is not that of a function.

(Thus, a circle is not the graph of a function).

We see that we can draw a vertical line, for example the dotted line in the drawing, which cuts the circle more than once.

Therefore the circle is not a function.

**Example 1. **Does this graph pass the vertical line test?

** **

**Solution.** It passes the vertical line test. Therefore, it is the graph of a function.

**Example 2.** Does this graph pass the vertical line test?

**Solution**. It does not pass the vertical line test because the vertical line we have drawn cuts the graph twice, so it is not the graph of a function.

A **one to one** function is a function which associates distinct arguments with distinct values; that is, every unique argument produces a unique result. It is not necessary for all elements in a co-domain to be mapped. A one to one function is also said to be an injective function.

A function *f* that is not injective is sometimes called **many-to-one**.

**Definition**. Let * * be a function whose domain is a set *X*. The function *f* is injective if

Similarly, if

- Put . Solve equation. If it yields then function is one-one; otherwise not.
- Alternatively, draw plot of the given function and apply the
*horizontal line test*. That is, draw a line parallel to*x*-axis such that it intersects as many points on the plot as possible. If it intersects the graph only at one point, then the function is one-one. - Alternatively, a function is a one-one function, if
*f(x)*is a continuous function and is either increasing or decreasing in the given domain.

**Exercise 1.** Let a function be given by:

Decide whether *f* is injective.

**Solution**. Applying the horizontal line test, draw a line parallel to *x-*axis to intersect the plot of the function as many times as possible.

We find that all lines drawn parallel to *x*-axis intersect the plot only once. Hence, the function is one-one.

**Exercise 2**. Let a function be given by:

Decide whether *f* is injective.

**Solution**. Draw the plot of the function and see intersection of a line parallel to *x*-axis.

We observe that there is no line parallel to *x*-axis which intersects the functions more than once. Hence, function is one-one.

**Exercise 3.** Let a function * * be given by:

^{
}

Decide whether *f* is injective.

**Solution**. We can solve and see whether to decide the function type.

Then

That is

We see * _{ }*that is not exclusively equal to . Hence, given function is not a one-one function, but a many – one function.

The conclusion is further emphasized by the intersection of a line parallel to *x*-axis, which intersects function plot at two points.

**Exercise 4**. Let a function * *be given by:

Decide whether *f* is injective.

**Solution**. The given function is a rational function. We have to determine function type.

We evaluate function for . If equation yields multiple values of *x*, then function is not one-one. Then,

Now,

We see that . It means pre-images are not related to distinct images. Thus, we conclude that function is not one-one, but many-one.

A function is an **onto** function if its range is equal to its co-domain. Onto functions are alternatively called **surjective** functions.

**Definition**. Let be a function whose domain is a set *X*. The function *f* is an onto function if and only if for every *y* in the co-domain* Y* there is at least one *x* in the domain *X* such that

**Exercise 5.** Let a function be given by:

Decide whether* f *is an onto function.

**Solution**. We can apply the definition to verify if f is onto.

Choosing for example , we obtain

* *is not onto.

**Definition**. A function is a **bijection** if the function is both one-one and onto and has the property that every element *y* *∈* *Y.* corresponds to exactly one element .

**Definition**. Let . Then, if it exists, the **inverse** of ƒ is the function , defined by the following rule:

Stated otherwise, a function is **invertible** if and only if its inverse relation is a function, in which case the inverse relation is the inverse function: the inverse relation is the relation obtained by switching *x* and* y* everywhere.

Not all functions have an inverse. For this rule to be applicable, each element must correspond to exactly one element *y ∈ Y* .

Most functions encountered in elementary calculus do not have an inverse.

**Definition**. A function admits an inverse function if the function is a bijection.

**Exercise 6.** Let a function * *be given by :

Decide whether has the inverse function and construct it.

**Solution**. The function is both one-one and onto, so the function f has the inverse function^{ }.

We construct an inverse rule in step-wise manner:

**Step 1**: Write down the rule of the given function .

Let the given rule be given by :

Let us put . Then

This relation gives us one value of image. For example, if , then

*Step 2*: Solve for :

Replace *x* which now represents image by the symbol * *and replace *y* which now represents independent variable by *x*. For the given function, , the new inverse rule is:

**Exercise 7**. Let a function be given by:

Decide whether* * has the inverse function and construct it.

**Solution**. The function is not one-one, so the function does not have the inverse function .

**Exercise 8. **Let a function* * be given by :

Decide whether has the inverse function and construct it.

**Solution**. The function * *is not one-one, so the function f does not have the inverse function ^{ }.

We can understand composition in terms of two functions.

Let there be two functions denoted as :

Observe that set *B* is common to two functions. The rules of the functions are given by *f (x)* and *g (x)* respectively. Our objective here is to define a new function and its rule.

Thinking in terms of relation, *A* and *B* are the domain and codomain of the function f. It means that every element *x* of A has an image *f (x)* in B.

Similarly, thinking in terms of relation, *B* and *C* are the domain and codomain of the function *g*.

In this function, *f (x) *which was the image of pre-image *x *in *A* is now pre-image for the function *g*. There is a corresponding unique image in set “*C*“. Following the symbolic notation, *f (x)* has image denoted by “*g(f (x))* ” in “C”. The gure here depicts the relationship among three sets via two functions (relations) and the combination function.

Function composition is a special relation between sets not common to two functions.

For every element *x* in *A*, there exists an element *f (x)* in set *B*. This is the requirement of function *f *by definition. For every

element *f (x)* in *B*, there exists an element *g(f (x))* in set* B*. This is the requirement of function *g* by definition. It follows, then, that for every element *x* in *A*, there exists an

element *g(f(x))* in set *C*. This concluding statement is definition of a new function :

By convention, we call this new function as * *and is read “*g* composed with *f*“.

The two symbolical representations are equivalent.

**Exercise 9. **Let two functions and be defined as follow:

Determine and .

**Solution**. According to definition,

Again, according to definition,

Importantly note that

It indicates that composition of functions is not commutative.

Let there be two functions denoted as :

The range of *f* is a subset of its co-domain *B*. But, set *B* is the domain of function *g *such that there exists image * g (f (x))* in C for every x in *A*. This means

that range of *f* is subset of domain of *g* :

Range of* f* ⊂ Domain of *g*

Clearly, if this condition is met, then composition * *exists. Following this conclusion, will exist, if

Range of* g* ⊂ Domain of *f*

And, if both conditions are met simultaneously, then we can conclude that both * * and g exist.

This is usually possible when all sets involved are sets of real numbers.

**Exercise 10**. Let two functions be defined as follows:

Check whether and exit for the given functions?

**Solution**. Here we have,

Domain of * *={1,2,3,4}

Range of* * ={2,3,4,5}

Domain of ={2,3,4,5}

Range of ={4,2,3,1} = {1,2,3,4}

Hence

Range of* f* ⊂ Domain of *g*

and

Range of* g* ⊂ Domain of *g*

This means that both compositions * * and * *exist for the given sets.

*1*. Functions and their graph. One–one and onto functions. Composite and inverse functions.

*2*. Linear inequalities. Systems of linear inequalities

*3*. Polynomial inequalities. Rational inequalities. Absolute-value inequalities

*6*. Differentiation. Derivative rules, the chain rule.

*7*. Application of differentiation: L'Hospital's Rule

*8*. Vertical, Horizontal and Slant asymptotes

*9*. Higher Order Derivatives. Applications of differentiation: local and absolute extremes of a function

*10*. Curve Sketching

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