Math · Calculus I · Worked example

Make a piecewise function continuous

Find the value of k that makes f continuous at x = 2, where f(x) = x² + k for x < 2 and f(x) = 3x − 1 for x ≥ 2.

f (x) = {\begin{cases} x^{2} + k, & x < 2 \\ 3 x - 1, & x \geq 2 \end{cases}

Find the value at 2

x = 2 belongs to the second piece, so f(2) = 3(2) − 1 = 5.

3 \cdot 2 - 1 = 5

For x > 2 the function is 3x − 1, which approaches 5.

lim_{x \to 2^{+}} (3 x - 1) = 5

For x < 2 the function is x² + k, which approaches 4 + k.

lim_{x \to 2^{-}} (x^{2} + k) = 4 + k

Continuity needs 4 + k = 5, so k = 1. Then both one-sided limits equal f(2) = 5.

\begin{array}{l} 4 + k = 5 \\ k = 1 \end{array}

k = 1.

Find a so that g(x) = ax + 1 for x < 1 and g(x) = x² + 3 for x ≥ 1 is continuous.

Show the answer and explanation

g(1) = 1 + 3 = 4, and the left piece approaches a + 1, so continuity needs a + 1 = 4.

\begin{array}{l} a + 1 = 4 \\ a = 3 \end{array}

In Graph, the two pieces meet at (2, 5). Change x² + 1 to x² + 3 and the left piece ends at height 7 instead: a jump discontinuity.

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