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- Closed interval : Again for same 2 real numbers, if x can take values between a and b , including a & b, then its a closed interval. i.e. a x b x [ ,a b] {square brackets are used}
- Half Open Interval : It contains both type of intervals, open, closed interval & closed open interval. In this type only one end point is included. a < x b x ( ,a b] a x < b x [ ,a b)
- Infinite intervals : Before going to intervals let us discuss first about infinity, denoted by . By infinity we mean that it is a very big real number, larger than any real number but how large, it is not fixed. When we say x R, we indirectly mean ă <x < or x ( ă , ) coming to infinite intervals now, whenever is at one or both the end points we never include them; i.e. ă <x < or x (ă , ) round brackets Not square brackets ă <x < a or x ( ă , a) ă <x a or x ( ă , a] x a or x ( ă , a]
So m e B a s i c De f i n i t i o n s Ć Domain : For a given function y = f (x), the set of values which x can take provided that for those values y is well defined, is known as Domain of the function.
for ex. y^ ^1 x, here x can take all real values except 0 because at x= 0 the value of y is invalid. Ć Range : For a given function y = f (x), the set of values which y can take, corresponding to each real number in the domain, is known as Range of function. for ex. y = x^2 , here x can take all real values but ycan take only positive values. Domain & Range can also be expressed as
Domain The values which input can take
Range The values which output can take
function
Ć Periodicity : A function is said to be periodic if it repeats itself after a certain interval. for example
We just covered the basic definitions of these terms, though their properties will be discussed later on in detail.
Cl a s s i f ic a t i o n o f f u n c t i o n s : (I) Algebraic functions : Functions consisting of finite number of terms involving powers and roots of independent variable with the operations +, ă , ï, are called algebraic functions.
for example : ^ ^2
, 1, 1 1 f x x x x x x
- Po l y n o m i a l f u n c t i o n s f (x ) = a 0 + a 1 x + a 2 x^2 + ... anxn , where a 0 , a 1 ... an R (i.e. real constants) and n is a non negative integer, is said to be a polynomial function of degree n (given an ^ 0) for ex. f (x) = 3x^3 + 2x^2 + x + 1 (polynomial of degree 3) f (x) = 1 (polynomial with degree 0)
f x x 3 x^2 1 (not a polynomial function)
(a) Constant Function If the range of a function f consists of only one number than f is called a constant function i.e. y = f (x) = a ex. y = f (x) = 1
(b) Identity Function The function y = f (x ) =x is known as identify function.
DOMAIN x R
RANGE y {a}
y = a f(x) = a
y
x DOMAIN x R
RANGE y R
y
x
y = x
45 Ĉ
This graph is repeating after every integeral interval.
-2 -1 1 2 3
T Y PE (I I ) EX PON EN T I A L & L OGA RI T H A M I C FU N CT I ON
- Ex p o n e n t i a l Fu n c t i o n The function g = f (x) = ax,^ a^ > 0,^ a^ ^ 1 is said to be an exponential function. It is divided into 2 parts depending on the value of a. for 0 < a < 1, y decreases as xincreases a > 1, y increases as x increases
Shape of curve as we can see from the graph that the value of y approaches zero but is never 0 (i.e. asymptote) and can take all positive values. Domain : x R Range : y (0, )
- L o g a r i t m i c Fu n c t i o n The function y = f (x) = loga x is known as logarithmic function. provided that x > 0 a > 0 and a 1 So the domain is very clear from the constraints only Domain : x (0, ) Range : y ( ă, ) Here also the function depends on the value of a.
1
0 < a < 1
Y
X a > 1
X
Y
1
O 0 < a < 1
Y
1 X
a > 0
Y
1
O (^) X
Pr o p e r t i e s o f l o g a r i t h m i c f u n c t i o n s :
- loge (ab) = loge a + log (^) e b
- l^ oge^ ^ ba = log (^) e b ă loge a
- loge am^ = m loge a
- loga a = 1
- logbm a = (^1) m logb a
- logb a =
1 loga b
- logb a =
log log
m m
a b
- al^ og^ ab^ = b
- al^ og^ bc^ = c l^ og^ ba
- If log (^) a x > loga y
, if 1 , if 0 1
x y a x y a
^
^ ^
- loga x = y x= ay
- loga x > y
, if 1 , if 0 1
y y
x a a x a a
(^) (^)
- logax < y
, if 1 , if 0 1
y y
x a a x a a
(^) (^) Also, when we say log x = y, then we take log with base 10. Similarly for loge x = y we write it as ln x = y (log with base e is also called natural log)
Illustration 2
Find domain of f (x) = ln (ă 2 + 3x ăx^2 ) Solution : for f (x) to be valid the log function should be valid and for that ă x^2 + 3x ă 2 > 0 Now, ă x^2 + 3x ă 2 > 0 x^2 ă 3 x + 2 < 0 x^2 ă 2 x ă x + 2 < 0 (x ă 2) (x ă 1) < 0 x (1, 2) domain (1, 2)
1 2
Now solving the second part 2 10 log 5 0 4
(^) x x
(^5 2 ) x 4 x 10
5 2 x 4 x 1
5 x ă x^2 4 x^2 ă 5 x + 4 0 (x ă 4) (x ă 1) 0 x [1, 4] ................ (ii) since both conditions have to be satisfied, we have to take the intersection of (i) & (ii) from (i) & (ii) x [1, 4] Domain [1, 4] (b) f (x) = log 10 (log 10 log 10 log 10 x) for function to be valid log 10 (log 10 log 10 x) > 0 & x > 0 log 10 (log 10 x) > 10^0 &^ x^ > 0 log 10 (log 10 x) > 1 &^ x^ > 0 log 10 x > 10^1 & x > 0 x > 10^10 & x > 0 combining both we get Domain (10^10 , )
Illustration 5
Find the domain of the function :
10
f x (^) log 1 x x
Solution : This question is a mix of algebraic & logarithmic functions.
Now,
10
f x (^) log 1 x x
(^1 )
We will solve both the parts separately & then combine their results to get the final results. log 10 (1 ă x) is valid when x < 1
1 log 1 x is valid when 1^ ă^ x^ ^ 1 (^ log^ a1 = 0) &^ x^ < 1 x < 1 except x = 0 (from 1 ă x 1 because at x = 0, denominator becomes 0) x (ă 1) ă {0} ................ (i) now solving the algebraic part for x 1 to be valid^ x^ + 1^ ^0 x ă 1 ................ (ii) combining (i) & (ii) we get x [ă 1, 0) (0, 1) The domain of the given functions is [ă 1, 0) (0, 1)
T r i g o n o m e t r y /c i rc u l a r f u n c t i o n s : Functions involving trigonometric ratios are called trigonometric functions.
(a) y = f x( ) = sin x
Domain : ( ă , ); Range : [ă 1, 1] ; Period : 2 ; Nature : odd ;
Interval in which the inverse can be obtained : ^ 2 , 2
(b) y = f x( ) = cos x
Domain : ( ă , ); Range : [ă 1, 1] ; Period : 2 ; Nature : even; Interval in which the inverse can be obtained : [0, ]
Y
X
(0, 1)
(0, 0) (0, ă1)
ă 3 ă 2 ă^2 3
Y
(^2) X
ă 1
1
/
ă 2 ă
ă /
(f) y = f x( ) = cosec x
Domain : R ă n, n (^) I; Range : (ă , ă1] (^) [1, ); Period : 2; Nature : odd ;
Interval in which the inverse can be obtained : ^ 2 , 0^ ^ 0, 2
I n v e r s e t r i g o n o m e t r i c /i n v e r s e c i r c u l a r f u n c t i o n s : Functions involving inverse of trigonometric ratios are called inverse trigonometric or inverse circular functions. (a) y = f x( ) = sină^1 x Domain : [ă 1, 1]
Range : ^ 2 , 2 Nature : odd ;
(b) y = f x( ) = cosă^1 x Domain : [ă 1, 1] Range : [0, ] Nature : neither even nor odd
(c) y = f x( ) = tană^1 x Domain : ( ă , ),
Range : ^ 2 , 2 Nature : odd;
Y
(^2) X 3 (^02)
1 ă 1
2
1
y = sină^1 x
ă 1
y
x
/
/
1 y = cosă^1 x
ă 1
y
x
/
0
x
y (^) /
/
0 ă
(d) y = f x( ) = cosecă^1 x Domain : ( ă , ă 1] (^) [1, )
Range : ^ ^2 , 0^ ^ ^ 0, 2 Nature : odd;
(e) y = f x( ) = secă^1 x Domain : ( ă , ă 1] (^) [1, )
Range : ^ 0,^2 ^ ^ ^2 , Nature : neither even nor odd
(f) y = f x( ) = cotă^1 x Domain : ( ă , ), Range : (0, ) Nature : neither even or odd.
Illustration 6
Find the domain for the following :
(a) f ( x ) cos sinx (b) ^ ^
1 2 y f x (^) sin ^ log 2 x 2 ^ ^ ^
Solution :
(a) f x( ) cos sinx is defined if
value under root is non-negative i.e. cos (sin x) 0 but we know that sin x lies between ă 1 & 1 ă 1 sin x 1 & for [ă 1, 1] cosine function is always + ve
1 x / y = cosecă^1 x
y
ă 1
/
ă
0
y = sec ă^1 x
x
/
ă 1 0 1
y
/
y
0 y = cot ă^1
x x
Solution :
(a) Given f (x) = log 10 sin (x ă 3) + (^16) x^2 for f x() to be well defined sin (x ă 3) > 0 & we know sin x is positive in (0, ). So generalising it, sin x will be positive in (2n + 0, 2 n + ) as 2 is the period of sin x. 2 n + 0 < (x ă 3) < 2n + 2 n + 3 < x < (2 n + 1) + 3 .......... (i) for the under root part 16 ă x^2 0 x^2 ă 16 0 (x ă 4) (x + 4) 0 x [ă 4, 4] ................ (ii) combining (i) & (ii) from (i) x (2 n + 3, (2n + 1) wheren = 0, 1, 2 (ii) x ^ [ă 4, 4]
The common region is (ă 2 + 3, ă + 3) (3, 4] Domain is (ă 2 + 3, ă + 3) (3, 4]
(b) given y^ cos^ ^1 2 2 sinx
^
for y to be defined.
1 2 1 (^2) sin x
solving first ^1 ^2 ^2 sin x
for (^) n= ă 1
ă 4 +
n = 0
denominator here can never be negative or zero because sin x [ ă 1, 1], cross multiplying ă (2 + sin x) 2 ă 2 ăsin x 2 sin x ă 3 which is true for all values of x as min. value of sin x is ă 1. ............. (i) now solving the second part (^2 ) 2 sin x 2 2 + sin x sin x 0 2 n + 0 x 2 n + 2 n x (2n + 1) ............... (ii) combining (i) & (ii) Domain (2 n, (2n + 1) )
So m e o t h e r f u n c t i o n s
- Absolute Value / Modulus function By absolute / modulus function we mean only the numerical value of the function, irrespective of its sign, from origin. This concept is analogous to distance. We can also say that modulus function is distance with respect to origin. Though, modulus function is defined as f : R R, f(x) = |x| here, Domain : x R Range : [0, ) Period : Non periodic Nature : even
Y
X
y = –x (^) y = x
O
Solution :
(a) We have
x 4 we can see that x 4
(^2 ) x (^) 4 & 2^2
a^ a b b
(^)
2 > |x ă 4| {we can do this because mod function is always positive} |x ă 4| < 2 ă 2 < x ă 4 < 2 2 < x < 6 x (2, 6) ă {4} (Note : Remember to remove 4 from domain students generally miss this step)
(b) we have,
x x
let |x| = y
yy^ ^12 ^0 y 2 or y 1 Note : we cannot include y = 2. |x| > 2 or |x| 1 (x > 2 or x <ă 2) or (ă 1 x 1) Domain (ă , ă 2) [ă 1, 1] (2, )
(c) we have |x ă 1| + |x ă 2| 4 we will solve this by finding the critical points & checking for values greater or smaller about these critical points. Here critical points are 1 & 2
|x ă 1| =
1 , 1 1 , 1
x x x x
(^)
2 2 ,^2 2 , 2 x x^ x x x ^ ^ (^) we can divide the values in 3 region i.e. < 1, between 1 & 2, & greater than 2.
1 2
Case 1 : when ă <x< 1 i.e. in this region |x ă 1| = ă (x ă 1) & |x ă 2| = ă (x ă 2) |x ă 1| + |x ă 2| 4 ă (x ă 1) ă (x ă 2) 4 ă 2 x + 3 4 2 x ă 1 0 x ă ó ............. (i) Case 2 : when 1 x 2 here in this region |x ă 1| = x+ 1 & |x ă 2| = ă (x ă 2) |x ă 1| + |x ă 2| 4 x ă 1 ă (x ă 2) 4 1 4 no solution for this solution ............. (ii) Case 3 : when x > 2 here both are positive i.e. |x ă 1| = x ă 1 & |x ă 2| = x ă 2 |x ă 1| + |x ă 2| 4 x ă 1 + x ă 2 4 2 x ă 3 4 x 7/2 ............. (iii) combining (i), (ii) & (iii)
Domain , 1 7 , 2 2 ^ ^ ^ (^)
(d)
x x x
x x x
x x x x
x x
................ (i) we know that x (ă2) now 2 cases arise for |x + 3|
