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

To find the behavior of the composition of functions \( (g \circ f)(x) \), we need to first understand the individual behaviors of \( f(x) \) and \( g(x) \).

Function Definitions

• \(f(x) = \begin{cases} x + 1, & \quad x < 0 \\ |x - 1|, & \quad x \geq 0 \end{cases}\) 
• \(g(x) = \begin{cases} x+1, & \quad x < 0 \\ 1, & \quad x \geq 0 \end{cases}\)

Finding \( f(x) \)

Let us examine \( f(x) \):

• For \( x < 0 \), \( f(x) = x + 1 \).
• For \( x \geq 0 \), \( f(x) = |x - 1| \), which evaluates to:
  • For \( x = 0, |0-1| = 1 \)
  • For \( x = 1, |1-1| = 0 \)
  • For \( x > 1, |x-1| = x - 1 \)

Finding \( g(f(x)) \)

Now, evaluate \( g(f(x)) \) by plugging \( f(x) \) into \( g(x) \):

• For \( x < 0 \), \( f(x) = x + 1 \). Since \( x + 1 < 0 \), \( g(x+1) = x + 2 \).
• For \( x = 0 \), \( f(0) = 1 \), \( g(1) = 1 \).
• For \( x = 1 \), \( f(1) = 0 \), \( g(0) = 1 \).
• For \( x > 1 \), \( f(x) = x - 1 \).
  • Since \( x - 1 \geq 0 \), \( g(x-1) = 1 \).

Behavior of \( (g \circ f)(x) \)

We need to analyze the continuity and differentiability of \( g(f(x)) \):

• Continuous at all points implies no jump discontinuities. Since both \( f(x) \) and \( g(x) \) are piecewise continuous, \( g(f(x)) \) is continuous everywhere.
• For differentiability, note that generally, a function composed with absolute values like \( |x| \) or piecewise defined as in \( f(x) \), can introduce points of non-differentiability. Here, this occurs because at \( x = 1 \), the composition changes form, having a potential sharp turn or cusp.

Conclusion

Thus, from the analysis, we confirm:

• \( (g \circ f)(x) \) is continuous everywhere.
• It is not differentiable exactly at one point, which is x = 1, where there is a change in derivative behavior due to the absolute value involved.

Thus, the correct answer is: Continuous everywhere but not differentiable exactly at one point.