The absolute value (or modulus) | a | of a real number a is the numerical value of a without regard to its sign. So,for example,the absolute value of 2 is 2,and the absolute value of –2 is also 2. The absolute value of a number may be thought of as its distance from zero.
From the above definition we can say that the absolute value of a is always either positive or zero, but never negative. If x is greater than or equal to zero, then use |x| = x , i.e., if a number is non negative, then its absolute value is itself. Whereas , If x is less than zero, then use |x| = - x i.e. if a number is negative, then its absolute value is its opposite.
The function f(x) = |x| is called the absolute value function. The domain of the absolute value function is the set of real numbers and the range is the set of positive real numbers.
In GeoGebra , absolute value function is written as abs(function name). You can enter the following functions in the input box of the applet and see the nature of graph.
1. f(x) = 3 abs(x-2) + 1 2.f(x) = -2 abs(x+2) + 3 3.f(x) = abs(x-1) 4.f(x) = 0.5 abs(x+1) - 2 5.f(x) = 3 - 2 abs(x-1)
Try to find the domain and range in each of the above cases with the movement of points B and C.
For any real number a the absolute
value or modulus of a is denoted
by | a | and is defined as
From the above definition we can say that the absolute value of a is always either positive or zero, but never negative. If x is greater than or equal to zero, then use |x| = x , i.e., if a number is non negative, then its absolute value is itself. Whereas , If x is less than zero, then use |x| = - x i.e. if a number is negative, then its absolute value is its opposite.
The function f(x) = |x| is called the absolute value function. The domain of the absolute value function is the set of real numbers and the range is the set of positive real numbers.
In GeoGebra , absolute value function is written as abs(function name). You can enter the following functions in the input box of the applet and see the nature of graph.
1. f(x) = 3 abs(x-2) + 1 2.f(x) = -2 abs(x+2) + 3 3.f(x) = abs(x-1) 4.f(x) = 0.5 abs(x+1) - 2 5.f(x) = 3 - 2 abs(x-1)
Try to find the domain and range in each of the above cases with the movement of points B and C.
Dear Sanjay,
ReplyDeleteIn fact you can input any function in the input box!