📝 Continuity of polynomial and rational functions (14 MCQs)
📖 From Calculus • 2. Limits and Continuity an Introduction • 14 questions available
What is Continuity of polynomial and rational functions?
Definition:
Polynomial functions are continuous everywhere on because their limits at any point equal their function values. Rational functions are continuous at all points where the denominator ; they are discontinuous only at zeros of the denominator, where vertical asymptotes or holes may occur, but on their domains, they are continuous.
Example:
Find discontinuities of .
Solution: makes denominator zero; at , there is a hole (removable discontinuity), so continuous elsewhere.
Reason:
This property assures that these common functions are continuous in most practical domains, simplifying limit evaluation and making them safe for use in modeling without worrying about unexpected breaks in standard operations.
📝 All Continuity of polynomial and rational functions MCQs
Q1. According to Theorem 1.5.4 (a), a polynomial is continuous ...
📖 Explanation: The theorem explicitly declares that any polynomial function has no breaks, jumps, or holes at any real number. Since a polynomial is defined for all real inputs and its limit equals its value everywhere, the function is continuous on the entire real line, i.e., everywhere.
Q2. For the rational function \y=\\frac{x^{2}-9}{x^{2}-5x+6}\, at which x‑values does the graph have a discontinuity?
📖 Explanation: The denominator \x^{2}-5x+6\ factors as \(x-2)(x-3)\. Zeroes of the denominator cause the rational function to be undefined, producing discontinuities at \x=2\ and \x=3\. Thus both points are discontinuities.
Q3. Which statement correctly describes the continuity of \f(x)=x^{2}\ and \g(x)=\\frac{1}{x-1}\?
📖 Explanation: The polynomial \x^{2}\ is continuous for all real \x\. The rational function \\\frac{1}{x-1}\ is continuous wherever its denominator is non‑zero, so it fails only at \x=1\. Hence the second statement is accurate.
Q4. Is the absolute‑value function \|x|\ continuous at \x=0\?
📖 Explanation: The piecewise definition \|x|=x\ for \x>0\ and \-x\ for \x<0\ yields limits of 0 from both sides. Since \|0|=0\, the two‑sided limit equals the function value, confirming continuity at \x=0\.
Q5. Let \f(x)=\\frac{x^{2}-4}{x-2}\ for \x\\neq2\ and \f(2)=5\. What type of discontinuity occurs at \x=2\?
📖 Explanation: Simplifying \\\frac{x^{2}-4}{x-2}=x+2\ gives a limit of 4 as \x\\to2\. Because the function is defined as 5 at \x=2\, the limit and the function value differ, creating a jump (point) discontinuity.
Q6. Compare \p(x)=x^{3}\ and \r(x)=\\frac{x^{3}-27}{x-3}\. Which statement is true?
📖 Explanation: The numerator of \r\ factors as \(x-3)(x^{2}+3x+9)\; canceling \(x-3)\ leaves \x^{2}+3x+9\. The limit as \x\\to3\ exists (27), but \r\ is undefined at 3, giving a removable discontinuity.
Q7. For \f(x)=\\frac{x^{2}-1}{x-1}\ ( \x\\neq1\ ), what is the continuity status at \x=1\ after simplifying?
📖 Explanation: Cancelling the factor \(x-1)\ yields \f(x)=x+1\ for \x\\neq1\. The limit as \x\\to1\ is 2, yet the original function is undefined at 1, so the point is a removable discontinuity.
Q8. Define \g(x)=\\frac{x^{2}-4}{x-2}\ for \x\\neq2\ and \g(2)=4\. Classify the continuity at \x=2\.
📖 Explanation: The simplified expression \x+2\ approaches 4 as \x\\to2\. Because the function is explicitly defined as 4 at \x=2\, the limit equals the function value, restoring continuity at that point.
Q9. Consider \h(x)=\\frac{|x|}{x}\ for \x\\neq0\. What is the behavior of the limit as \x\\to0\ and the continuity at 0?
📖 Explanation: For \x>0\, \h(x)=1\; for \x<0\, \h(x)=-1\. The right‑hand limit is 1 and the left‑hand limit is -1, so the two‑sided limit does not exist, making the function discontinuous at \x=0\.
Q10. For \f(x)=\\frac{x^{2}-9}{x-3}\, which statement correctly describes continuity at \x=3\?
📖 Explanation: Factorizing gives \(x-3)(x+3)\ in the numerator; cancelling \(x-3)\ leaves \x+3\. The limit at \x=3\ exists (6), but the original function is undefined there, resulting in a removable discontinuity.
Q11. For \F(x)=\\begin{cases}\\frac{x^{2}-4}{x-2},&x\\neq2\\\\k,&x=2\\end{cases}\, which value of \k\ makes \F\ continuous at \x=2\?
📖 Explanation: The limit as \x\\to2\ of \\\frac{x^{2}-4}{x-2}=x+2\ equals 4. Setting \k=4\ aligns the function value with the limit, eliminating the hole and ensuring continuity at \x=2\.
Q12. Let \p(x)=\\frac{x^{4}-16}{x^{2}-4}\. Identify all points of discontinuity and their type.
📖 Explanation: Both numerator and denominator factor as \(x^{2}-4)(x^{2}+4)\ and \(x-2)(x+2)\ respectively, cancelling the common factor \(x^{2}-4)\. The simplified function \x^{2}+4\ is defined everywhere except at \x=\\pm2\, where the original expression had holes—removable discontinuities.
Q13. Given \g(x)=\\frac{1}{(x-1)(x-2)}\, which statement about its continuity is correct?
📖 Explanation: A rational function is continuous wherever its denominator is non‑zero. Here the denominator vanishes at \x=1\ and \x=2\, creating two points of discontinuity. Everywhere else the function is defined and its limit equals its value, so it is continuous on \\\mathbb{R}\\setminus\\{1,2\\}\.
Q14. Define \h(x)=\\frac{x^{2}}{|x|}\ for \x\\neq0\ and set \h(0)=0\. What is the continuity status at \x=0\?
📖 Explanation: For \x>0\, \h(x)=x\; for \x<0\, \h(x)=-x\. Both one‑sided limits approach 0 as \x\\to0\. Since the function is defined as 0 at \x=0\, the limit equals the function value, establishing continuity at that point.