Continuity, question 26

Topic: Continuity
Difficulty: Capstone
Suggested time: 10 min
Marks: 5
Theory: UBC Math 100 §2.3 CLP-1 §1.6Key results

Find the values of the constants $a$ and $b$ that make $f$ continuous everywhere. Justify your answer using the definition of continuity.

f(x) = \begin{cases} \dfrac{x^{3} + 1}{x + 1}, & x < -1 \\[10pt] ax + b, & -1 \leq x \leq 2 \\[6pt] \dfrac{x^{2} + 5x - 14}{x - 2}, & x > 2 \end{cases}

Write your reasoning in full, then compare it against the model solution.

Your answer

Theory UBC Math 100 §2.3 CLP-1 §1.6Key results

Worked solution

Strategy

Away from the joins every piece is continuous, so continuity everywhere comes down to two conditions, one at each join. Find the outer one-sided limits by factoring, set each equal to the value of the middle piece, and solve the pair of equations.

Working

On x < −1 the denominator x + 1 is never zero, on x > 2 the denominator x − 2 is never zero, and the middle piece is a polynomial. So f is continuous everywhere except possibly at x = −1 and x = 2.
$\lim_{x \to -1^{-}} \frac{x^{3} + 1}{x + 1} = \lim_{x \to -1^{-}} \bigl(x^{2} - x + 1\bigr) = 1 + 1 + 1 = 3$
Sum of cubes: x³ + 1 = (x + 1)(x² − x + 1). Cancel, then substitute.
$\lim_{x \to 2^{+}} \frac{x^{2} + 5x - 14}{x - 2} = \lim_{x \to 2^{+}} (x + 7) = 9$
Factor as (x + 7)(x − 2) and cancel x − 2.
$x = -1:\ -a + b = 3, \qquad x = 2:\ 2a + b = 9$
At each join the outer one-sided limit must equal the value of the middle piece, which is also its limit from the other side.
$3a = 6 \quad\Longrightarrow\quad a = 2, \quad b = 5$
Subtract the first equation from the second, then substitute back.
$\lim_{x \to -1} f(x) = 3 = f(-1), \qquad \lim_{x \to 2} f(x) = 9 = f(2)$
With a = 2 and b = 5 both joins meet the definition: the limit exists and equals the value. The middle piece is the line through (−1, 3) and (2, 9).

Answer

a = 2, \qquad b = 5

What earns the marks

Explains that each piece is continuous on its own open interval (neither outer denominator is zero there), so only x = −1 and x = 2 need checking.
Finds the left-hand limit at −1 by factoring x³ + 1 = (x + 1)(x² − x + 1) and cancelling: the limit is 3.
Finds the right-hand limit at 2 by factoring x² + 5x − 14 = (x + 7)(x − 2) and cancelling: the limit is 9.
Uses the middle piece for the values f(−1) = −a + b and f(2) = 2a + b, and sets each equal to the matching one-sided limit.
Solves −a + b = 3 and 2a + b = 9 to get a = 2 and b = 5, and concludes that at each join the limit exists and equals the value.

Check

At x = −1.01 and x = 2.01 the outer pieces give 3.0301 and 9.01, close to 2x + 5 at the joins, 3 and 9.

Where this goes wrong

Substituting x = −1 into (x³ + 1)/(x + 1), getting 0/0 and treating the left side as 0. The outer pieces never reach their joins; what matters is their one-sided limits, found after factoring.

Leans on

Limits by factoring and cancelling, from Limits

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Continuity practice question 26