Choose a Constant to Make a Piecewise Function Continuous at the Join

For continuity at x = c, the left-hand limit, right-hand limit and defined value f(c) must all agree. Match the branches at the joining point, then check the value assigned there.

Three checks, not just two formulas

Continuity at a point requires a defined function value, an existing two-sided limit, and equality between that limit and the value. A piecewise formula may use one rule to the left, another to the right, and a separate value exactly at the join. All three parts matter.

Polynomial branches are continuous within their own intervals, so their one-sided limits at a finite join can be found by substitution. For more complicated branches, calculate the relevant limit rather than assuming that direct substitution is allowed.

Worked example

Define f(x) = kx + 1 when x < 2, and f(x) = x² − 1 when x ≥ 2. Find k so that f is continuous at 2. From the left, the limiting value is 2k + 1. From the right, it is 2² − 1 = 3. The defined value f(2) also comes from the second branch and equals 3.

Match the left limit to 3: 2k + 1 = 3, so k = 1. With that choice, the left branch approaches 3, the right branch approaches 3, and f(2) = 3. The continuity conditions are all satisfied.

Continuous does not always mean smooth

With k = 1, the left branch’s slope is 1, while the right branch’s slope at 2 is 4. The function is continuous there but not differentiable there. Matching the output heights removes a gap; matching derivative values would be an additional requirement.

Similarly, |x| is continuous at zero but has unequal one-sided slopes. Do not use the word “smooth” as a substitute for the three continuity conditions. The question may ask only for continuity, in which case no derivative calculation is necessary.

A separately assigned value

If h(x) = x + 2 for x ≠ 1 and h(1) = a, the two-sided limit at 1 is 3. Continuity requires a = 3. Changing one function value does not change the nearby limit, but it can determine whether the function is continuous at that exact point.

Common mistake

Read the inequality signs in each branch carefully. The branch containing equality supplies f(c). If the one-sided limits are unequal, changing only f(c) cannot repair the jump. If they agree, a separately chosen value can fill the missing point. State both the constant and the final common value as a check.

Join par left, right aur actual value—teeno same hone chahiye.

Three practice questions with worked answers

1. A function equals ax for x < 1 and 3x − 1 for x ≥ 1. Find a for continuity at 1.

   The left limit is a and the right limit/value is 2. Therefore a = 2, with common value 2.

2. Let g(x) = x² + 1 for x ≠ 2 and g(2) = b. Find b for continuity.

   The nearby polynomial limit is 2² + 1 = 5. Set b = 5 so that the defined value matches that limit.

3. The left limit is 4 and right limit is 7. Can any value at the join make the function continuous?

   No. The two-sided limit does not exist because the one-sided limits differ. Reassigning the value at one point cannot make 4 equal 7.

Related study guides

• Limits and Derivatives Class 11: Standard Limits and Solved Examples
• Continuity and Differentiability Class 12: Key Results and Solved Examples

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Continuity. No textbook or exam-board affiliation is claimed.