Concept: Function & Graphs
CONTENTS
Functions and Graphs is a very important chapter from CAT and XAT perspective. You can expect 2 to 3 quesetions from this chapter in CAT. SNAP and NMAT do not ask questions from this chapter.
A function y = f(x) is where x is the input and f(x) is the output.
x is the independent variable here and y is the dependent variable.
Example: f(x) = 2x + 3 is a function whose input is x and (2x + 3) is the output.
If we have to calculate value of f(2), it means we need to calculate the output when input i.e., x = 2.
∴ f(2) = 2 × 2 + 3 = 7.
Similarly, if y = f(x) = x2 - 2x + 3, then
f(-1) = (-1)2 - 2 × (-1) + 3 = 1 + 2 + 3 = 6.
A function is defined only if for every input (x) there is only one output (f(x) or y).
y = 3x3 + 3x + 5 is a function, but
y2 = 2x + 3 is not a function.
because at x = 3, y can be +3 or -3 i.e., there are 2 outputs possible for the same input.
The composition of a function is an operation where two functions say f and g generate a new function say h in such a way that h(x) = g(f(x)). It means here function g is applied to the function of x. So, basically, a function is applied to the result of another function.
Solution:
f(x) = 2x + 3
⇒ f(g(x)) = 2 × g(x) + 3
⇒ f(g(x)) = 2 × x2 + 3
Solution:
g(x) = x2 + 1
⇒ g(1) = 12 + 1 = 2
Now, f(x) = 3x + 1
⇒ f(g(1)) = f(2) = 3 × 2 + 1 = 7
Note: f(g(x)) can also be written as fog(x). This means f or g(x).
Solution:
f(x) = 3x - 1
⇒ f(1) = 3 × 1 - 1 = 2
Now, fof(1) = f(2) = 3 × 2 - 1 = 5.
Now, fofof(1) = f(5) = 3 × 5 - 1 = 14.
Solution:
f(x) = 3x - 1
⇒ f(0) = 3 × 0 - 1 = -1
Now, gof(0) = g(-1) = (-1)2 = 1.
Now, fogof(0) = f(1) = 3 × 1 - 1 = 2.
A function which has different definitions depending upon the value of the independent variable is called a Piecewise Function. Thus, the definition of the function changes according to the input.
Example:
Here, if x is ≥ 0 f(x) = x and
if x < 0 , f(x) = -x
∴ f(2) = 2 but f(-2) = -(-2) = 2
A periodic function is a function that repeats itself after regular intervals. Thus, if f(x) = f(x + c), where c is some constant, for all values of x, then f(x) is a periodic function. c is called the period/periodicity of the function.
The most common periodic functions are the trigonometric functions like sin x, cos x etc.
Solution:
𝑔(𝑥) = 𝑔(𝑥 + 1)+ 𝑔(𝑥 −1)
⇒ g(x + 1) = g(x) - g(x - 1)
Let g(1) = a and g(2) = b
⇒ g(3) = g(2) - g(1) = b - a
⇒ g(4) = g(3) - g(2) = b - a - b = -a
⇒ g(5) = g(4) - g(3) = -a - (b - a) = -b
⇒ g(6) = g(5) - g(4) = -b - (-a) = a - b
⇒ g(7) = g(6) - g(5) = a - b - (-b) = a
⇒ g(8) = g(7) - g(6) = a - (a - b) = b
We can see that g(7) = g(1), g(8) = g(2) and so on
∴ g(x + 6) = g(x)
g(x) is a periodic function whose periodicity is 6.
A function f(x) is an even if f(-x) = f(x)
Exmaple: f(x) = 22
Here, f(-x) = (-x)2 = x2 = f(x)
A function f(x) is a odd if f(-x) = -f(x)
Exmaple: f(x) = 23
Here, f(-x) = (-x)3 = -x3 = -f(x)
Note: Not all functions are even or odd. Some function may neither be even nor odd.
If f(x) = y, then inverse function is defined as f-1y = x.
In a function y = f(x), x is the input and y is the output,
In an inverse function x = f-1y, y is the input and x is the output.
Example: If f(x) = y = 2x + 3, then
⇒ x =
⇒ f-1y = g(y) =
We can replace y with x, and hence the inverse function g(x) =
Not all functions have an inverse. Whether or not an inverse exists for a function, can be found using graphs. If the reflection of the graph of a function about the line y = x satisfies the vertical line test, then the inverse exists.