Math Problem Statement

Which of the following function is not continuous everywhere? (A) f(x) = -x^2 if x ≤ 0 and f(x) = x^2 if x > 0 (B) f(x) = 2x + 1 if x ≤ 0 and f(x) = -2x + 1 if x > 0 (C) f(x) = 1 if x ≤ 0 and f(x) = e^x if x > 0 (D) f(x) = sin(x) if x ≤ 0 and f(x) = cos(x) if x > 0

Solution

Let’s analyze each function for continuity at x=0x = 0, as that is the point where the definition of the functions changes. A function is continuous at x=0x = 0 if:

  1. The left-hand limit (LHLLHL) and right-hand limit (RHLRHL) exist and are equal at x=0x = 0.
  2. The value of the function at x=0x = 0 matches these limits.

(A) ( f(x) = \begin{cases}

-x^2 & \text{if } x \leq 0 \ x^2 & \text{if } x > 0 \end{cases} )

  • Left-hand limit as x0x \to 0^-: LHL=x20LHL = -x^2 \to 0.
  • Right-hand limit as x0+x \to 0^+: RHL=x20RHL = x^2 \to 0.
  • Value at x=0x = 0: f(0)=x2=0f(0) = -x^2 = 0.

Since LHL=RHL=f(0)LHL = RHL = f(0), f(x)f(x) is continuous at x=0x = 0.


(B) ( f(x) = \begin{cases}

2x + 1 & \text{if } x \leq 0 \ -2x + 1 & \text{if } x > 0 \end{cases} )

  • Left-hand limit as x0x \to 0^-: LHL=2(0)+1=1LHL = 2(0) + 1 = 1.
  • Right-hand limit as x0+x \to 0^+: RHL=2(0)+1=1RHL = -2(0) + 1 = 1.
  • Value at x=0x = 0: f(0)=2(0)+1=1f(0) = 2(0) + 1 = 1.

Since LHL=RHL=f(0)LHL = RHL = f(0), f(x)f(x) is continuous at x=0x = 0.


(C) ( f(x) = \begin{cases}

1 & \text{if } x \leq 0 \ e^x & \text{if } x > 0 \end{cases} )

  • Left-hand limit as x0x \to 0^-: LHL=1LHL = 1.
  • Right-hand limit as x0+x \to 0^+: RHL=e0=1RHL = e^0 = 1.
  • Value at x=0x = 0: f(0)=1f(0) = 1.

Since LHL=RHL=f(0)LHL = RHL = f(0), f(x)f(x) is continuous at x=0x = 0.


(D) ( f(x) = \begin{cases}

\sin(x) & \text{if } x \leq 0 \ \cos(x) & \text{if } x > 0 \end{cases} )

  • Left-hand limit as x0x \to 0^-: LHL=sin(0)=0LHL = \sin(0) = 0.
  • Right-hand limit as x0+x \to 0^+: RHL=cos(0)=1RHL = \cos(0) = 1.
  • Value at x=0x = 0: f(0)=sin(0)=0f(0) = \sin(0) = 0.

Since LHLRHLLHL \neq RHL, f(x)f(x) is not continuous at x=0x = 0.


Conclusion

The function (D) is not continuous everywhere.

Would you like detailed steps or further clarifications?


Relative Questions:

  1. What are the formal conditions for a function to be continuous at a point?
  2. Can a function be continuous everywhere but not differentiable everywhere? Provide examples.
  3. How can we determine continuity at endpoints of a domain?
  4. What happens if only LHL=RHLLHL = RHL but f(0)f(0) is undefined?
  5. How is piecewise function continuity related to real-world applications?

Tip:

Always check both the limits and the actual function value to ensure continuity.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Continuity of functions
Limits and their evaluation
Piecewise functions

Formulas

Left-hand limit (LHL): lim x→c^- f(x)
Right-hand limit (RHL): lim x→c^+ f(x)

Theorems

A function f(x) is continuous at x = c if LHL = RHL = f(c).

Suitable Grade Level

Grades 11-12