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Question

Question: The function \[f\] is defined by \[f(x) = \left\\{ 1 - x,x < 0 \\\ 1,x = 0 \\\ x + ...

The function ff is defined by

1 - x,x < 0 \\\ 1,x = 0 \\\ x + 1,x > 0 \\\ \right.$$ . Draw the graph of $$f(x)$$ .
Explanation

Solution

A function is nothing but an expression which defines a relationship between the independent variable and the dependent variable. For instance, let us consider a function f(x)=yf(x) = y , here yy is the dependent variable and xx is the independent variable. The variable yy will get a value for each value of xx .

Complete step by step answer:
It is given that f(x) = \left\\{ 1 - x,x < 0 \\\ 1,x = 0 \\\ x + 1,x > 0 \\\ \right.
that is,
When x<0x < 0 , f(x)=1xf(x) = 1 - x
When x=0x = 0 , f(x)=1f(x) = 1
When x>0x > 0 , f(x)=x+1f(x) = x + 1
First, we have to find the points from the given function.
It is given that, when x<0x < 0 , f(x)=1xf(x) = 1 - x , so xx can take values like x=1,2,3,4,...x = - 1, - 2, - 3, - 4,... .
Let’s substitute the values of xx in the function.
For x=1x = - 1 , f(1)=1(1)=1+1=2f( - 1) = 1 - ( - 1) = 1 + 1 = 2 . So, the point is (1,2)( - 1,2) .
For x=2x = - 2 , f(2)=1(2)=1+2=3f( - 2) = 1 - ( - 2) = 1 + 2 = 3 . So, the point is (2,3)( - 2,3) .
For x=3x = - 3 , f(3)=1(3)=1+3=4f( - 3) = 1 - ( - 3) = 1 + 3 = 4 . So, the point is (3,4)( - 3,4) .
For x=4x = - 4 , f(4)=1(4)=1+4=5f( - 4) = 1 - ( - 4) = 1 + 4 = 5 . So, the point is (4,5)( - 4,5) .
We can keep on finding the point since there is no limit for the xx value, so we will stop here. Let us find the points for the next condition.
It is given that, when x=0x = 0 , f(x)=1f(x) = 1 . Here xx has only one value that is 0, so f(0)=1f(0) = 1 . So, the point is (0,1)(0,1) .
Now let us find the points for the next condition.
It is given that, when x>0x > 0 , f(x)=x+1f(x) = x + 1 , so xx can take values like x=1,2,3,4,...x = 1,2,3,4,... .
Let’s substitute the values of xx in the function.
For x=1x = 1 , f(1)=1+1=2f(1) = 1 + 1 = 2 . So, the point is (1,2)(1,2) .
For x=2x = 2 , f(2)=2+1=3f(2) = 2 + 1 = 3 . So, the point is (2,3)(2,3) .
For x=3x = 3 , f(3)=3+1=4f(3) = 3 + 1 = 4 . So, the point is (3,4)(3,4) .
For x=4x = 4 , f(4)=4+1=5f(4) = 4 + 1 = 5 . So, the point is (4,5)(4,5) .
Now let’s plot these points in the graph with the scale xx axis 1unit=1cm1unit = 1cm and yy axis 1unit=1cm1unit = 1cm and join the points.

This is the required graph for the given function.

Note: Since the condition of the given problem doesn’t have any limit point, we have found some points to plot it in the graph. Point to remember while drawing graph: Scale of the graph is important; it has to be uniform. The given function doesn’t stop anywhere since xx has indefinite values, we will be getting a value of yy for each value of xx .