Question

Let f and g be two function defined by \(f(x) =   \begin{cases}     x+1       & \quad x<0\\    |x-1|,   & \quad x \geq0   \end{cases}\) and g(x) = \(f(n) =   \begin{cases}     x+1,       & \quad x<0 \\     1,  & \quad x\geq0   \end{cases}\) Then (gof)(x) is 

• A. Continuous everywhere but not differentiable at x = 1
• B. Continuous everywhere but not differentiable exactly at one point
• C. not continuous at x = – 1
• D. differentiable everywhere

Answer 1

The Correct Option is B

Given Functions: 

\[ f(x) = \begin{cases} x + 1, & x < 0 \\ 1 - x, & 0 \leq x < 1 \\ x - 1, & x \geq 1 \end{cases} \]

\[ g(x) = \begin{cases} x + 1, & x < 0 \\ 1, & x \geq 0 \end{cases} \]

Step 1: Compute \( g(f(x)) \)

Case 1: When \( x < 0 \)

For \( x < 0 \), \( f(x) = x + 1 \). Substituting \( f(x) \) into \( g(x) \):

\[ g(f(x)) = g(x + 1) = (x + 1) + 1 = x + 2. \]

Case 2: When \( 0 \leq x < 1 \)

For \( 0 \leq x < 1 \), \( f(x) = 1 - x \). Substituting \( f(x) \) into \( g(x) \):

\[ g(f(x)) = g(1 - x). \]

Since \( 1 - x \geq 0 \) for \( 0 \leq x < 1 \), \( g(1 - x) = 1 \).

Case 3: When \( x \geq 1 \)

For \( x \geq 1 \), \( f(x) = x - 1 \). Substituting \( f(x) \) into \( g(x) \):

\[ g(f(x)) = g(x - 1). \]

Since \( x - 1 \geq 0 \) for \( x \geq 1 \), \( g(x - 1) = 1 \).

Step 2: Final Composition of \( g(f(x)) \)

Combining all cases:

\[ g(f(x)) = \begin{cases} x + 2, & x < 0 \\ 1, & x \geq 0 \end{cases} \]

Step 3: Analyze Continuity and Differentiability

Continuity:

The function \( g(f(x)) \) is continuous everywhere because there are no jumps or breaks in its definition.

Differentiability:

• For \( x < 0 \), \( g(f(x)) = x + 2 \). The derivative is: \[ \frac{d}{dx}(x + 2) = 1. \]
• For \( x \geq 0 \), \( g(f(x)) = 1 \). The derivative is: \[ \frac{d}{dx}(1) = 0. \]
• At \( x = 0 \): The left-hand derivative is 1, and the right-hand derivative is 0. Since these are not equal, \( g(f(x)) \) is not differentiable at \( x = 0 \).

Final Observations:

• \( g(f(x)) \) is continuous everywhere.
• \( g(f(x)) \) is not differentiable at \( x = 0 \).
• \( g(f(x)) \) is differentiable everywhere else.