Introduction to the Concept of Functions

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Let us understand the concept of functions through some examples:

  1. The area of a circle can be expressed in terms of its radius \(A = \pi {r^2}\). The area A is dependent on the radius r. In the language of functions, we say that A is a function of r.

  2. The volume V of a sphere is a function of its radius. The dependence of V on r is given by \(V = \frac {4}{3}\pi {r^3}\).

  3. The acceleration a of a body of fixed mass m is a function of the force F applied on the body: \(a = \frac {F}{m}\).

  4. The power P dissipated in a resistor of fixed resistance R is a function of the current I passing through the resistor: \(P = {I^2}R\).

  5. Suppose that a taxi charges a down payment of Rs. 100, and subsequently, Rs. 10 for every kilometer travelled. The taxi fare F is a function of the distance d travelled, and the dependence between the two quantities is given by \(F = 100 + 10d\) (F is in rupees and d is in kilometers).

Whenever we say that a variable quantity y is a function of a variable quantity x,we mean to say that: y depends on x; the value of y is determined by the value of x. We can write this dependence as follows:

\[y = f\left( x \right)\]

Observe this notation carefully. It says that: y is a function of x, or depends on x. Do not make the mistake of interpreting the right side as a product (of f and x)! The term \(f\left(x \right)\) implies a function of x– in other words, a quantity or value which depends on the value of x.

Suppose that y is a function of x. The exact dependence between y and x is that is twice of the square of x. We can write this as follows:

\[y = f\left( x \right) = 2{x^2}\]

The first half of this relation,\(y = f\left( x \right)\) tells us that is a function of x, or y depends on x. The second half, \(f\left( x \right) = 2{x^2}\), tells us the exact definition of this function, i.e., how exactly the function’s output value will be calculated based on the input value. For example,

  • When \(x = 1\), then \(y = f\left( 1 \right) = 2{\left( 1 \right)^2} = 2\)

  • When \(x =  - 2\), then \(y = f\left( { - 2} \right) = 2{\left( { - 2} \right)^2} = 8\)

  • \(f\left( 3 \right) = 2{\left( 3 \right)^2} = 18\)

  • \(f\left( { - \frac{1}{2}} \right) = 2{\left( { - \frac{1}{2}} \right)^2} = \frac{1}{2}\)

Thus, for any value of the input variable (which is x), we can calculate the value of the output variable(which is y), through the relation \(y =f\left( x \right) = 2{x^2}\).

Example 1:  is a function of x, and the function definition is given as follows:

\[y = f\left( x \right) = \frac{1}{{1 + {x^2}}}\]

Find the output values of the function for \(x = 0\), \(x =  - 1\) and \(x = \sqrt 2 \).

Solution: We have:

\[\begin{align}&f\left( 0 \right) = \frac{1}{{1 +{{\left( 0 \right)}^2}}} = \frac{1}{1} = 1\\&f\left( { - 1} \right) = \frac{1}{{1+ {{\left( { - 1} \right)}^2}}} = \frac{1}{2}\\&f\left( {\sqrt 2 } \right) =\frac{1}{{1 + {{\left( {\sqrt 2 } \right)}^2}}} = \frac{1}{{1 + 2}} =\frac{1}{3}\end{align}\]

Example 2: The length of a rectangle is twice that of its breadth. Express the area of the rectangle as a function of its (i) length (ii) diagonal length.

Solution: Consider the following figure:

Area of rectangle - example 1

(i) In terms of the length l,the area A is

\[A = f\left( l \right) = l \times \frac {l}{2} =\frac {{{l^2}}}{2}\]

(ii) The diagonal length d is given be

\[{d^2} = {l^2} + \frac{{{l^2}}}{4} =\frac{{5{l^2}}}{4}\]

Thus,

\[l = \frac{{2d}}{{\sqrt 5 }},\,\,\,b = \frac{l}{2} =\frac{d}{{\sqrt 5 }}\]

Now, the area A can be expressed as a function of d, as follows:

\[A = g\left( d \right) = lb = \left({\frac {{2d}}{{\sqrt 5 }}} \right)\left( {\frac{d}{{\sqrt 5 }}} \right) =\frac {{2{d^2}}}{5}\]

Note that when expressing A as a function of l, we used the letter f to represent the function.When expressing A as a function of d, we used the letter g.This is because the function definitions are different in the two cases.

Example 3: A rectangular storage container with an open top has a volume of 10 m3. The length of its base is twice its width. Material for the base costs Rs 10 per square meter; material for the sides costs Rs 6  per square meter. Express the cost of materials as a function of the width of the base.

Solution: The following diagram depicts an open container with length and width of the base being 2w and respectively, and its height being h:

Volume of rectangular container - example 1

The area of the base is \(2 w \times w = 2{w^2}\). Thus, the cost in rupees of the material for the base is.

\[10 \times 2{w^2} = 20{w^2}\]

Two of the sides have area wh  and the other two have area 2wh, so the cost of the material for the sides is.

\[6\left\{ {2\left( {wh} \right) + 2\left( {2wh}\right)} \right\} = 36wh\]

The total cost is therefore

\[C = 20{w^2} + 36 wh\]

To express C as a function of alone, we need to eliminate h, and we do so by using the fact that the volume of the container is 10 m3. Thus,

\[\begin{array}{l}\left( {2 w} \right)\left( w\right)\left( h \right) = 10\\ \Rightarrow \,\,\,h =\frac{5}{{{w^2}}}\end{array}\]

Substituting this into the expression for C, we have

\[\begin{align}&C = 20{w^2} + 36 wh\\\,\,\,\,\, &\quad= 20{w^2} + 36 w\left({\frac{5}{{{w^2}}}} \right)\\\,\,\,\,\, &\;\;\;= 20{w^2} +\frac {{180}}{w}\end{align}\]

This expression gives C as a function of w.

Example 4: A cone has a variable height h and a variable base radius r, but the sum of h and r is fixed. The cone is made of a material of density \(\rho \). Express the mass of cone as a function of its height h.

Solution: The volume V of a cone is given by:

\[V = \frac{1}{3}\pi {r^2}h\]

Let the (fixed) sum of h and be k. Thus, \(r = k - h\), and so:

\[V = \frac{1}{3}\pi h{\left( {k - h} \right)^2}\]

The mass of the cone can now we expressed as a function of h; m will be \(\rho \) times the volume V:

\[m = f\left( h \right) = \rho V = \frac{1}{3}\pi\rho h{\left( {k - h} \right)^2}\]

This is the required function definition.