Find a and b for Continuity of a Piecewise Function —Example-7| Differential Calculus

Limits and Continuity · Worked Example-7

Finding a and b for continuity of a piecewise function

Matching left-hand and right-hand limits at x = 1 and x = 2 to solve for the unknown constants.

ax + b, x ≤ 1 x², 1 < x ≤ 2 bx + a, x > 2

We want the graph to have no breaks or jumps at the two joining points, \(x = 1\) and \(x = 2\). Recall the continuity condition at a point \(x = c\):

\[ \lim_{x\to c^{-}} f(x) = \lim_{x\to c^{+}} f(x) = f(c) \]

\[ f(x)= \begin{cases} ax+b, & x \leq 1 \\ x^2, & 1 < x \leq 2 \\ bx+a, & x > 2 \end{cases} \]

Step 1

Continuity at x = 1

Compare the left-hand limit, right-hand limit, and the function value at \(x = 1\):

Left-hand limit

\(\displaystyle\lim_{x\to 1^{-}} (ax+b) = a+b\)

Right-hand limit

\(\displaystyle\lim_{x\to 1^{+}} x^2 = 1\)

f(1)

\(a(1)+b = a+b\)

Equating all three:

\[ a + b = 1 \quad \cdots (1) \]

Step 2

Continuity at x = 2

Compare the left-hand limit, right-hand limit, and the function value at \(x = 2\):

Left-hand limit

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

Right-hand limit

\(\displaystyle\lim_{x\to 2^{+}} (bx+a) = 2b+a\)

f(2)

\((2)^2 = 4\)

Equating all three:

\[ 2b + a = 4 \quad \implies \quad a + 2b = 4 \quad \cdots (2) \]

Step 3

Solve the simultaneous equations

Subtract equation (1) from equation (2) to eliminate \(a\):

\[ \begin{aligned} (a+2b) - (a+b) &= 4 - 1 \\ b &= 3 \end{aligned} \]

Substitute \(b = 3\) back into equation (1):

\[ \begin{aligned} a + 3 &= 1 \\ a &= -2 \end{aligned} \]

\( a = -2, \quad b = 3 \)

Summary

• Value of a−2
• Value of b3
• Continuous atx = 1 and x = 2

Watch it worked out

A step-by-step video walkthrough of this same problem.

Engineering Mathematica · Linear Algebra Notes · MA25C01

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