Math Problem Statement

Which of the following functions is not continuous on the interval \(-\infty < x < \infty\)? Options: (A) \(f(x) = 4x^2 - 2x + 1\), (B) \(g(x) = \frac{1}{x^3 + 3x^2 - 2x - 5}\), (C) \(h(x) = \cos(\pi x)\), (D) \(k(x) = \frac{1}{e^x}\).

Solution

To determine which function is not continuous on the interval <x<-\infty < x < \infty, let's examine each function for continuity.

Option Analysis

  1. Option A: f(x)=4x22x+1f(x) = 4x^2 - 2x + 1

    • This is a polynomial function, and polynomial functions are continuous on the entire real line. Therefore, f(x)f(x) is continuous on (,)(-\infty, \infty).
  2. Option B: g(x)=1x3+3x22x5g(x) = \frac{1}{x^3 + 3x^2 - 2x - 5}

    • This function is a rational function, which means it will be discontinuous at values of xx that make the denominator zero.
    • To check for points of discontinuity, we need to find the roots of the denominator x3+3x22x5=0x^3 + 3x^2 - 2x - 5 = 0. Solving this for xx would yield specific values where the function is undefined, leading to discontinuity at those points.
    • Since there are values of xx that make the denominator zero, g(x)g(x) is not continuous over the entire real line.
  3. Option C: h(x)=cos(πx)h(x) = \cos(\pi x)

    • This is a composition of a trigonometric function with a linear function, both of which are continuous everywhere on the real line.
    • Therefore, h(x)h(x) is continuous on (,)(-\infty, \infty).
  4. Option D: k(x)=1exk(x) = \frac{1}{e^x}

    • This is an exponential function in the denominator, and ex0e^x \neq 0 for any real xx. Thus, 1ex\frac{1}{e^x} is defined and continuous for all real values of xx.
    • Hence, k(x)k(x) is continuous on (,)(-\infty, \infty).

Conclusion

The function g(x)=1x3+3x22x5g(x) = \frac{1}{x^3 + 3x^2 - 2x - 5} (Option B) is not continuous on (,)(-\infty, \infty) because the denominator can be zero for certain values of xx, causing discontinuities.

Would you like further details on solving for the roots of the denominator in Option B, or do you have any other questions?

Related Questions

  1. How do you find points of discontinuity in rational functions?
  2. What methods can you use to solve a polynomial equation like x3+3x22x5=0x^3 + 3x^2 - 2x - 5 = 0?
  3. Why are trigonometric functions generally continuous on the real line?
  4. Can exponential functions in the denominator ever lead to discontinuities?
  5. How does the degree of a polynomial affect its continuity?

Tip

For any rational function f(x)=p(x)q(x)f(x) = \frac{p(x)}{q(x)}, check for continuity by identifying values of xx that make q(x)=0q(x) = 0, as these points lead to discontinuities in the function.

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Math Problem Analysis

Mathematical Concepts

Continuity of Functions
Rational Functions
Polynomials
Trigonometric Functions
Exponential Functions

Formulas

Continuity of Rational Functions: \( \frac{p(x)}{q(x)} \) is discontinuous where \( q(x) = 0 \)

Theorems

Continuity Theorem for Polynomials
Continuity Theorem for Trigonometric Functions
Continuity Theorem for Exponential Functions

Suitable Grade Level

Grades 11-12