Continuity, question 27

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

Find the constants $a$ and $b$ that make $f$ continuous everywhere, and enter the product $ab$ . Here $\log$ is the natural logarithm.

f(x) = \begin{cases} \dfrac{e^{2x} - 1}{e^{x} - 1}, & x < 0 \\[10pt] ae^{x} + b, & 0 \leq x \leq \log 3 \\[6pt] e^{2x} + 1, & x > \log 3 \end{cases}

Your answer

An integer or a fraction.

How sure are you?

SureNot sureGuessing

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

Worked solution

Strategy

Two joins, two unknowns. Find the limit of each outer piece at its join, set it equal to the value of the middle piece there, and solve the pair of equations.

Working

For x < 0, eˣ < 1, so the denominator eˣ − 1 is never 0; the other two pieces are defined everywhere. So f is continuous except possibly at x = 0 and x = log 3.
$\lim_{x \to 0^{-}} \frac{e^{2x} - 1}{e^{x} - 1} = \lim_{x \to 0^{-}} \frac{(e^{x} - 1)(e^{x} + 1)}{e^{x} - 1} = \lim_{x \to 0^{-}} \bigl(e^{x} + 1\bigr) = 2$
e²ˣ − 1 is a difference of squares in eˣ. Cancel the common factor, then substitute.
$x = 0:\quad ae^{0} + b = a + b = 2$
The middle piece gives f(0), which must match the limit from the left.
$\lim_{x \to (\log 3)^{+}} \bigl(e^{2x} + 1\bigr) = e^{2\log 3} + 1 = 9 + 1 = 10$
e^{2 log 3} = (e^{log 3})² = 3² = 9.
$x = \log 3:\quad ae^{\log 3} + b = 3a + b = 10$
The middle piece gives f(log 3), which must match the limit from the right.
$2a = 8 \quad\Longrightarrow\quad a = 4, \quad b = -2, \quad ab = -8$
Subtract the first equation from the second, then substitute back.

Answer

a = 4, \quad b = -2, \quad ab = -8

Check

The outer pieces give about 1.990 at x = −0.01 and 10.0 at x = 1.1, close to 4eˣ − 2 at the joins, 2 and 10.

Where this goes wrong

Substituting x = 0 into the left piece and reading 0/0 as 0, or writing e^{2 log 3} as 6. The left piece never reaches 0, so its limit is what counts, found by factoring; and a multiple in the exponent becomes a power, so e^{2 log 3} = 9.