Differential Calculus: Function

FUNCTION

Calculus begins with the study of functions. Functions are very fundamental in mathematics. The term function was coined by Leibnitz in 1673. A function is a tool that scientists and mathematicians use to describe relationship between varying quantities

For example:

1. The speed of a rocket is a function of its payload.

2. The price of a ticket is a function of where you sit.

3. The perimeter of a circle is a function of its radius.

We now formally define a function.

Definition 2.1 Function

A function f from a set A to a set B is a rule that assigns to each element x ∈ A a unique element y in B.

 The element y in B is called the image of x under for the value of ƒ at x and is written symbolically as f(x).

Thus, we have y = f(x).

The set A is called the domain of the function f and B is called the codemain of f.

The set of all values of the function f is called the range of the function. It is denoted by f(A) or Rf and range is a subset of B.

Thus,

(1) In applications, usually a machine or a system is represented by a function f. So we can consider an element x in the domain of ƒ as input and its value f(x) in the range as output.

(2) Suppose ƒ is a function from A to B and y ∈ B such that y = f(x).

Then x is called the independent variable and y is called the dependent variable.

In other words, a variable in the domain is the independent variable and a variable in the range is the dependent variable.

Definition 2.2 Real Functions

If the domain and range of a function ƒ are subsets of the real number set ℝ, then ƒ is called a real function

We study here only real functions. By a function we always mean real function.

1. Methods of Representing a Function

There are four common methods of representing a function.

1. Algebraically (by an equation)

For example, area A of a circle depends on its radius r and is given by A = πr2. A is a function of r, since for each given r there is unique value for A.

2. Numerically (by table of values)

The table of values

represents a function, since for each x there is unique value for y.

3. Geometrically or visually (by graphs)

The graph represents a function, since for each value of x between 0 and 2 there is unique value for y given by the curve.

4. Verbally (by a description in words)

For example, Issac Newton's Law of Universal Gravitation is stated as below. The gravitational force of attraction between two bodies in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This is the verbal description of the formula

where G is gravitational constant and if m1, m2 are constant masses, then the force of attraction F is a function of r.

2. Arrow Diagram

A function f: A → B can also be represented by an arrow diagram. The assignments are shown by arrows in figure.

Examples

Test whether the following arrow diagrams define a function from A to B

Yes, since for each element in A, there is unique image in B.

No, since image of element 1 ∈ A is not unique

No, since the element 4 in A is not assigned to any element.

Definition 2.3 Graph of a given Function

If ƒ is a function with domain D, then the graph of ƒ is the set of ordered pairs {(x, f(x)) : x ∈ D, f(x) ∈ ƒ(D)}.

In other words, the graph of the function f consists of the set of points (x, y) in the xy-plane such that y = f(x), x is in the domain of f.

3. Domain and Range

Generally, a function is given by an expression or formula.

For example,

When the domain of a function ƒ is not stated explicitly, it is to be understood that the domain is the set of all real numbers x for which f(x) is real. This set is called the natural domain of the function f. In fact it is the largest possible domain.

If we want to restrict the domain for some reason, then it can be done. The range may also then change.

4. Intervals

The domain and range of a real function are expressed as an interval of real numbers.

Closed interval [a, b] = {x |a ≤ x ≤ b}

Open interval (a, b) = {x |a < x < b}

One side open interval [a, b) = {x |a ≤ x < b}

(a, b] = {x |a < x ≤ b}

x ≥ a ↔ [a,∞); x > a ↔ (a, ∞)

5. Solution of Quadratic Inequalities

1. If a > 0, consider the quadratic inequality ax2 + bx + c ≥ 0 or ≤ 0 If ax2 + bx + c = 0 has real and different roots α, β (α < β)

Then ax2 + bx + c = a(x − α)(x - β)

⸫ ax2 + bx + c ≥ 0 ⇒ (x − α) (x - β) ≥ 0 

⇒ x − α ≥ 0 and x - β ≥ 0 ⇒ x ≥ α and x ≥ β

⇒ x ≥ β

or x − α ≤ 0 and x - β ≤ 0 ⇒ x ≤ α and x ≤ β

⇒ x ≤ α

⸫ solutions of ax2 + bx + c ≥ 0 are x ≤ α or x ≥ β    (1)

i.e., the solutions lie outside the interval [α, β]

WORKED EXAMPLES

Find the domain of the following functions

Solution 

We shall denote the domain of ƒ as Dƒ.

Find the domain and range of the following functions

Solution

We have seen that given a function, its graph can be drawn in the xy-plane. But every graph in the xy-plane is not a function.

To decide whether a given graph represents a function or not, we have a simple test called the vertical line test.

6. The Vertical Line Test

A given curve in the xy plane is the graph of some function f, if no vertical line meets the curve more than once.

For example,

consider the two graphs given below

The graph in fig (i) represents a function because no vertical line meets the curve more than once.

The graph in fig (ii) does not represent a function because a vertical line meets the curve more than once.

Algebraically, at x = a, the image, which is the value of y, is not unique. Therefore it is not a function.

WORKED EXAMPLES

Example 18 

Consider the equation x2 + y2 = 4 

Discuss it represents a function or not.

Solution 

Given x2 + y2 = 4

Its graph is a circle with centre (0, 0) and radius 2.

The vertical line x = 1 meets the curve in 2 points.

So, it does not represent a function.

If we solve algebraically, we get two solutions

They represent the upper and lower semicircles and their union is the full circle.

Each of these satisfy the vertical line test and so they represent 2 functions

So the equation x2 + y2 = 4 represent 2 functions

This leads to the concept of piecewise-defined function or split-definition of function.

7. Absolute Value Function

Its graph is given by the equations y = x if x ≥ 0 and y = -x < 0

For example,

If x = -5, then f(-5) = -(-5) = 5

If x = 3/2, then f(3/2) = 3/2

The graph of absolute value function is shown here

From the graph we find |x| ≥ 0  ∀ x

Its domain is R = (-∞, + ∞) and range = [0, ∞)

Remark: If x ∈ R, √x2 = |x|

√4 = ± 2 is wrong, but √4 = 2

is wrong, but = |-2| = 2. Square root of a positive number is positive.

Greatest integer function (or floor function)

[x] = the greatest integer ≤ x.

⸫  [2.9] = 2, [−2.9] = −3

The graph resembles steps and so it is called a step function. Its domain is R = (-∞, ∞) but range is Z = {0, ±1, ±2, ....} [x] is also denoted by [x] and is called integer floor function in computer science and other areas.

WORKED EXMPLES

Example 19 

A function ƒ is defined by

Evaluate f(0), ƒ(1); ƒ(2), ƒ(−2) and sketch the graph

Solution Given

Since 0 and 1 lie in x ≤ 1, f(0) = 1 − 0 = 1, f(1) = 1−1 = 0 

Since 2 > 1, f(2) = 22 = 4, f(−2) = 1-(-2) = 3, since -2 < 1 

For x ≤ 1, the graph of the function is y = 1 − x, which is a straight line.

Put x = 0, then y = 1

Put y = 0, then x = 1  Two points on the line are (1, 0), (0, 1)

For x > 1, y = x2 is the parabola, which is a known curve. So we draw the graph. 

Example 20 

The graph of a function ƒ is given here

(i) Find ƒ(1) and f(5)

(ii) What are the domain and range of f?

Solution

(i) we note that the point (1, 3) lies on the graph

⸫ f(1) = 3

When x = 5, the graph is about 0.7 unit below x-axis

⸫ ƒ(5) = −0.7

(ii) The graph is drawn in the interval [0, 7]

⸫  the domain of the function is the closed interval [0, 7]

The values of y lie between −2 and 4

⸫  Range = [-2, 4].

Example 21 

Find the formula for the function given by the graph.

Solution 

The given graph is a triangle OAB, where 0(0, 0), A(1, 1), B(2, 0)

Example 22 

Find an expression for the function whose graph is the given curve

Solution 

The given graph consists of a semicircle with centre (0, 0) and radius = 2 and two line segments joining (2, 0), (4, 4) and (-2, 0), (-4, 4). 

The equation of circle is x2 + y2 = 4

Equation of upper semicircle is

Now the equation of the line joining the points (2, 0) and (4, 4) is

The equation of the line joining (−2, 0) and (-4, 4) is y − 0 = m (x + 2),

Example 23 

Find the expression for the function whose graph is the line segment joining the points (1,-3) and (5, 7)

Solution 

The graph of the line segment joining the point A (1, −3) and B (5, 7) is shown here Slope of AB =

Example 24 

Find the expression for the function whose graph is the bottom half of the parabola x + (y − 1)2 = 0.

Solution 

The given parabola is (y - 1)2 = -x [y2 = -4ax form]

Vertex is (0, 1) and axis of symmetry is y = 1 and the curve is towards the negative x-axis. The graph is the bottom half parabola

Example 25 

Find the domain and sketch the graph of the functions 

Solution

(a) Given g(x) =

The domain is the set of values of x for which g(x) is real

⇒ x – 5 ≥ 0 ⇒ x ≥ 5

⸫ Dg = [5, ∞)

To draw the graph:

This is a parabola with vertex (5, 0) and axis y = 0 is the x-axis and the curve is towards the positive x-axis.

Since y ≥ 0, the curve is the upper half parabola.

(b) Given F(x) = |2x + 1| 

F(x) is real for all x ∈ R

DF = R = (-∞, ∞)

(c) Given G(x) =

⸫ DG = R - {0}

⸫ the graph is y = 4 if x > 0

and y = 2 if x < 0

(d) Given g(x) =

⸫ Dg = R - {0}

Example 26 

For the following piece-wise defined functions find the domain and sketch the graph

Solution

To draw the line y = x + 2, we need two parts on the line. 

⸫ Put x = 0, then y = 2 and put y = 0, then x = -2

⸫ the points are (0, 2) and (−2, 0)

To draw the line y = 1 - x, we need two parts on the line.

Put x = 0, then y = 1 and put y = 0, then x = 1

⸫ the points are (0, 1), (1, 0)

The two lines are drawn as in figure.

(ii) f(x) =

⸫ Domain is Df = R = (−∞, ∞)

The graph is y = x + 9  if x < -3

y = -2x if -3 ≤ x ≤ 3

and y = -6 if x > 3

To draw the line y = x + 9, we need two points on the line.

⸫ put x = 0, then y = 9 and put y = 0, then x = -9

⸫ the two points are (0, 9), (−9, 0)

y = -2x passes through the origin. 

y = -6 is parallel to the x-axis

We shall now find functions from verbal problems.

Example 27 

An open box is to be made from a rectangular cardboard of size 16 × 30 meters by cutting off equal squares from the four corners and then bending up the sides. Find the volume as a function of the side x of the square.

Solution 

Given board is 16 × 30 meters

Equal squares are cut off from the corners

Let x meter be the side of the squares cut of from the corners. 

Then sides of the box will be 16 - 2x, 30 - 2x, x and x ≥ 0 

⸫ the volume of the box will be

V = (16 – 2x) (30 – 2x) x

we must have 16 - 2x ≥ 0 ⇒ x ≤ 8

and 30 - 2x ≥ 0 ⇒ x ≤ 15

The common values will be x ≤ 8

⸫ the volume function is V(x) = (16 – 2x) (30 – 2x) x, 0 ≤ x ≤ 8.

Example 28 

A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 30 m, express the area A of the window as a function of the width x of the window

Solution 

Let x be the width of the window Then radius of the semicircle is x/2

If y is the height of the window, then perimeter of the window (not including the base of semicircle) is

8. Even and Odd Function

Definition 2.4

A function f is said to be even if f(−x) = f(x) and odd if f(−x) = -f(x) for all x in its domain D.

Remarks: can an even or odd function exist in the interval [1, 3] ? 

No, if x ∈ [1, 3] then -x ∉ [1, 3] and so f(-x) is not defined on [1, 3]. 

Hence the definition can not be satisfied.

⸫ even or odd functions do not exist on [1, 3]

Hence the definition implies that for even or odd function to exist, the domain D should be an interval of the form (-a, a) or [-a, a]

Geometrically, the graph y = f(x) of an even function is symmetric about the y-axis, since for every point (x, y) on the graph the points (-x, y) also lie on the graph.

The graph y = f(x) of an odd function is symmetric about the origin, since for every point (x, y) on the graph the points (-x, -y) also lie on the graph.

WORKED EXAMPLES

Example 1 

Determine the following functions even, odd or neither

1. f(x) = x2 + x

2. g(x) = 1- x4

3. h(x) = 2x - x2

4. f(x) = x|x|

5. y = x - sin x

Solution 

Note that the domain of the functions from (1) to (5) is R = (-∞, ∞). So, we can discuss even or odd nature

Example 2 

Determine the domain of the function given by and test it is even or odd.

Solution 

Given f(x) =

We know log, x is defined if x > 0

Example 3 

Express the following functions in piece-wise form and draw their graph.

1. g(x) = 3 + |2x -5|

2. f(x) = |x| + |x -1|

Solution

1. Given g(x) = 3 + |2x -5|

By the definition of modulus function, we get

To draw the line y = 2x - 2, we need two points on the line

Put x = 0, then y = -2 and put y = 0, then x = 1

⸫ the points are (0,-2), (1, 0)

Similarly, the two points on the line y = 8 - 2x are (0, 8), (4, 0)

2. Given f(x) = |x| + |x -1|

The values x = 0 and x = 1 decide the rewriting of these modulae. 0 and 1 divide the real line into 3 parts.

So, we have 3 cases,

The graph of the function y = -2x + 1, y = 1 and y = 2x - 1, which are straight lines.

To draw the line y = 2x + 1, we need two points on the line.

⸫ put x = 0, then y = 1 and put y = 0, then x = 1/2

⸫ two points on the line are (0, 1) and ( 1/2, 0).

Differential Calculus 2-25

y = 1 is line parallel to the x-axis between x = 0 and x = 1 

To draw the y = 2x - 1, we need two points on the line

put x = 0, then y = -1 and y = 0, x = 1/2

⸫ two points on the line are (0, −1) and ( 1⁄2, 0)

Now, we draw the graph

The graph resembles a trough and so, it is called a trough function.

EXERCISE 2.1

Find the domain of the following functions

Find the formula for each function represented by the graphs below.

Find formula for the function in the following

12. An open rectangular box with volume 2 m3 has a square base. Express the surface area of the box as a function of the length of a side of the base.

13. The point (x, y) lies on the graph of the line 2x + 4y = 5. If L is the distance from (x, y) to the origin. Write L as a function of x.

14. The figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units.

(a) Express the y-coordinate of P as function of x.

(b) Express the area of the rectangle as a function of x.

Hint: A(1, 0), B(0, 1): Equation of AB is x + y = 1 ⇒ y = 1 − x 

⸫ y-coordinate of P is y = 1 - x, 0 < x < 1.

ANSWERS TO EXERCISE 2.1