Question

The function $f(x) = max \{(1 - x), (1 + x), 2\}, x \in (- \infty, \infty)$ is

• A. continuous at all points
• B. differentiable at all points
• C. differentiable at all points except at x = 1 and x = - 1
• D. continuous at all points except at x = 1 and x = -1, where it is discontinuous

Answer 1

The Correct Option is C

Answer (a) continuous at all pointsAnswer (c) differentiable at all points except at x = 1 and x = - 1.

Let's consider the function f(x)=max{(1−x),(1+x),2}, which can be defined as follows:

f(x)=⎩⎨⎧​1−x,2,1+x,​if x≤−1if −1≤x≤1if x≥1​

Hence, we can observe that:

• As x approaches -1 from the left (denoted as x→−1−), f(x) approaches 1−(−1)=21−(−1)=2.
• As x approaches -1 from the right (denoted as x→−1+), f(x) approaches 22.
• At x=−1, f(x)=2.

Thus, the left-hand limit, right-hand limit, and the value of f(x) at x=−1 are all equal, implying that f(x) is continuous at x=−1.

It's also evident that f(x) is continuous at x=1 as well.

Additionally, f(x) can be analyzed based on its piecewise structure: it is a polynomial function for x≤−1 and x≥1, and a constant function for −−1≤x≤1. Consequently, f(x) is continuous for all x.

Now, considering the derivatives at x=−1 and x=1:

• The left derivative of f(x) at x=−1 (Lf′(x)) is -1.
• The right derivative of f(x) at x=−1 (Rf′(x)) is 0.
• The left derivative of f(x) at x=1 (Lf′(x)) is 0.
• The right derivative of f(x) at x=1 (Rf′(x)) is 1.
• Hence, differentiable at all points except at x = 1 and x = - 1.