Examples of Functions
"f(x) = x^{2} " is the general manner to represent a function. It is said as f of x is equal to x square. This is represented as f = {(1,1), (2,4), (3,9)}. The domain and range of a function is given as D= {1, 2, 3}, R={1,4, 9}. Here is a representation of a function in math as an ordered pair.
Function in math
Function
NonFunction
Algebra of Functions
Important Notes
Example 1: Given two functions f and g as, f (x) = √(x  2) and g (x) = ln( 1 + x^{2}). Find the composite function (g ∘f )(x).
Given: functions f(x) and g(x).
(g ∘ f )( x ) = g ( f (x))
Since f (x) =√(x  2), substituting x = f (x), we get,
(g ( f (x)) = ln( 1+ f (x)2)
= ln( 1 +√(x  2) 2)
= ln(1 + ( x  2 ))
= ln((x  1))
∴ (g ∘ f )( x ) = ln(x  1)
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.
(i) In terms of the length l,the area A is
A = f( l ) = l × l/2 = l^{2}/2
(ii) The diagonal length d is given be
d^{2} = l^{2} + l^{2} /4 = 5l^{2}/4
Thus,
l =2d/ √5, b = l/2 =d/ √5
Now, the area A can be expressed as a function of d, as follows:
A = g( d) = lb = 2d/ √5 × d/ √5
=2d^{2}/5
Area as a function of length = l^{2}/2 and Area as a function of diagonal length = 2d^{2}/5
Example 3: 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 ρ. Express the mass m of the cone as a function of its height h.
The volume V of a cone is given by:
V = 1/3 π r^{2}h
Let the (fixed) sum of h and r be k. Thus, r = k  h, and so:
V = 1/3 π h(kh)^{2}
The mass of the cone can now we expressed as a function of h; m will be ρ times the volume V:
m = f(h) = ρ V
m = 1/3 π ρ h(kh)^{2}
m = 1/3 π ρ h(kh)^{2} is the required function definition.
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