Solved Examples of Modulus Function

It can be seen from the graph that the relationship between x and y for function y = |x| is as follows

• If x = -5 then y = |-5| = 5
• If x = -4 then y = |-4| = 4
• If x = -3 then y = |-3| = 3
• If x = -2 then y = |-2| = 2
• If x = -1 then y = |-1| = 1
• If x = 0 then y = |0| = 0
• If x = 1 then y = |1| = 1
• If x = 5 then y = |5| = 5
• If x = 10 then y = |10| = 10
• If x = 15 then y = |15| = 15

 

Thus it can be easily inferred from the graph that for all possible values of x, the function f(x) remains positive. 

Solved Examples On Modulus

Question 1:

Solve |x + 4| = 10 using modulus function.

Solution: 

As we have noted earlier, the modulus function always gives positive output. Therefore for the function |x + 4| = 10, there can be two situations 

If x + 4 > 0, then |x + 4| = x + 4 and

If x + 4 < 0, then |x + 4| = −(x + 4)

 

Situation 1: If x + 4 > 0

|x + 4| = x + 4

⇒ x + 4 = 10

⇒ x = 10 − 4 = 6

 

Situation 2: If x + 4 < 0, we have

|x + 4| = − (x + 4)

⇒ − (x + 4) = 10

⇒ − x − 4 = 10

⇒ x = − 4 − 10 = − 14

Therefore, the solution for x is −14 and 6.

It can thus be said that −14 < x < 6.

 

Question 2:

Solve x for |x2 – 7x + 12| using modulus function:

 

Solution:

|x2 – 7x + 12| = |(x – 3) (x – 4)| = |f(x)|

 

According to the definition of modulus function,

When f(x) is positive

|f(x)| = f(x)

When  f(x) is negative

| f(x) |= -f(x)

 

Thus, 

f(x) = (x – 3)(x – 4) is  non-negative or zero when x = (- ∞, 3] ∪ [4, ∞)

f(x) = (x – 3)(x – 4) is negative when x = (3, 4)

 

So, |x2 – 7x + 12| = (x2 – 7x + 12) when x = (-∞, 3] ∪ [4, ∞) and

|x2 – 7x + 12| = -(x2 – 7x + 12) when x = (3, 4)

 

Question 3:

If |x2 – 7x + 12| + |x2 – 9x + 20| = 0. Find x.

 

Solution:

We know that a modulus function is always a positive number. 

In the question, |x2 – 7x + 12| + |x2 – 9x + 20| = 0

Thus, when two different modulus functions add up to zero, then it must mean that both the modulus functions are also equal to zero (The sum of two positive numbers can never be zero unless they both are equal to 0).

Thus, 

x2 – 7x + 12 = 0 

(x – 3) (x – 4) = 0

x = 3, 4

Similarly, 

x2 – 9x + 20 = 0

(x – 4) (x – 5) = 0

x = 4, 5

 

Thus it can be seen that both the equations are zero at only one point, ie, when x = 4

Thus the only possible solution for x for |x2 – 7x + 12| + |x2 – 9x + 20| = 0 is x = 4

Conclusion

Thus in this article we have studied that the modulus of a real number x is given by the modulus function. The function provides a non-negative value for x, implying that the function’s output is always positive. We have also learnt how to plot the graph of the modulus function, which depicts the relationship between x and y.