Question

# Consider the function $f(x) = \max \{ 1,\left| {x - 1} \right|,\min \{ 4,\left| {3x - 1} \right|\} \} \forall x \in R$. Then find the value of f (3).

Hint: Recall the definition for min and max functions. Then substitute x as 3 in f (x) and evaluate the min and max functions to obtain the answer.

We have three special functions in f (x) which are min function, max function, and modulus function.
The min function or the minimum function takes in numbers as input and gives the minimum of the numbers as output. For example, min{1, 2} is 1 because 1 is less than 2 and it is the minimum of 1 and 2.
The max function or the maximum function takes in numbers as input and gives the maximum of the numbers as output. For example, max{1, 2} is 2 because 2 is greater than 1 and it is the maximum among the two.
Modulus function or the absolute value function always returns the absolute value of the number which is always positive. It is represented as $|x|$. For example, $| - 5|$ is equal to 5.
It is given that $f(x) = \max \{ 1,\left| {x - 1} \right|,\min \{ 4,\left| {3x - 1} \right|\} \} \forall x \in R$.
We need to determine f (3).
$f(3) = \max \{ 1,\left| {3 - 1} \right|,\min \{ 4,\left| {3(3) - 1} \right|\} \}$
Simplifying, we get:
$f(3) = \max \{ 1,\left| 2 \right|,\min \{ 4,\left| {9 - 1} \right|\} \}$
$f(3) = \max \{ 1,\left| 2 \right|,\min \{ 4,\left| 8 \right|\} \}$
We know that the modulus of 8 is 8.
$f(3) = \max \{ 1,\left| 2 \right|,\min \{ 4,8\} \}$
Now, the min{4, 8} is 4 because 4 is the minimum among 4 and 8.
$f(3) = \max \{ 1,\left| 2 \right|,4\}$
We know that the modulus of 2 is 2.
$f(3) = \max \{ 1,2,4\}$
The maximum among 1, 2, and 4 is 4.
$f(3) = 4$