Applications: Continuity

The bill is a function of the number of units, written in two pieces that meet at 200 units. Continuity at the join means the value the first piece reaches and the value the second piece starts from are the same number, and that one equation fixes the rate.

1. Below the join the first rule applies, so C(x) = 0.28x for 0 ≤ x ≤ 200. The limit from the left is lim_{x → 200⁻} C(x) = 0.28 × 200 = 56, that is $56.
2. Above the join the second rule applies, so the limit from the right is lim_{x → 200⁺} C(x) = a × 200 + 12 = 200a + 12 dollars.
3. Continuity at 200 units means the two one-sided limits agree with each other and with the value there, so 200a + 12 = 56.
4. Subtract 12 from both sides to get 200a = 44, then divide both sides by 200. (a) a = 0.22, so the rate above the band is $0.22 for each unit.
5. (b) A month of 350 units uses the second rule: C(350) = 0.22 × 350 + 12 = 77 + 12 = $89. Check: at 200 units the second rule gives 0.22 × 200 + 12 = $56, the same as the first rule, so the graph has no break.

answer(a) a = 0.22; (b) $89

techniqueContinuity at a Point

Continuity at a join asks three things of one point: the value there, the limit from the left and the limit from the right must all be the same number. The gradient is a separate question, and the two answers need not agree.

1. At the first join the left rule gives the value and the limit from the left: P(5) = 12 and lim_{x → 5⁻} P(x) = 12, because that rule is the constant 12.
2. The limit from the right uses the middle rule: lim_{x → 5⁺} P(x) = 12 − 0.8(5 − 5) = 12. The value and both one-sided limits are 12, so P is continuous at x = 5.
3. At the second join the middle rule gives lim_{x → 20⁻} P(x) = 12 − 0.8 × 15 = 12 − 12 = 0, and the last rule gives P(20) = 0 and lim_{x → 20⁺} P(x) = 0. (a) So P is continuous at both joins, at 12 kW and at 0 kW.
4. For part (b), the first rule is a constant, so the gradient just below x = 5 is 0. The middle rule is 12 − 0.8(x − 5), a straight line of gradient −0.8, so the gradient just above x = 5 is −0.8 kW for each degree.
5. (b) The gradient changes from 0 to −0.8 at the join, so the graph turns a corner at x = 5. The power itself does not change there, so the heater is continuous at x = 5 and not smooth: continuity asks that the graph have no break, not that it have no corner.

answer(a) P is continuous at both joins: the power is 12 kW at x = 5 and 0 kW at x = 20; (b) the gradient is 0 just below x = 5 and −0.8 kW for each degree just above, so the graph has a corner there and is continuous but not smooth

techniqueContinuity at a Point

A discontinuity is removable when the limit at the point is a number, even though the formula has no value there. Find that limit by canceling the factor that made the bottom zero, and store it as the missing value.

1. Put x = 2 into the formula: the top is 2³ − 8 = 0 and the bottom is 2 − 2 = 0. The form 00 is not a number, which is why the meter has nothing to show.
2. Factorize the top as a difference of two cubes: x³ − 8 = (x − 2)(x² + 2x + 4).
3. On the way to 2 the reading is near 2 and never equal to 2, so x − 2 is not zero and cancels: Q(x) = x² + 2x + 4 for every x ≠ 2.
4. That polynomial is continuous everywhere, so lim_{x → 2} Q(x) = 2² + 2 × 2 + 4 = 12. (a) Storing 12 liters per minute for x = 2 makes the reading continuous there, so the discontinuity is removable.
5. (b) At 2.5 volts, Q(2.5) = 2.5² + 2 × 2.5 + 4 = 6.25 + 5 + 4 = 15.25 liters per minute. Check against the original formula: 2.5³ − 82.5 − 2 = 7.6250.5 = 15.25.

answer(a) 12 liters per minute; (b) 15.25 liters per minute

techniqueTypes of Discontinuity · Continuity at a Point

Look at the top and the bottom of the fraction separately as the object nears the focal point. A fixed positive top over a bottom that shrinks to zero grows past every number, and a limit that is not a number cannot be matched by any value.

1. As u approaches 20 from above, the top tends to 20 × 20 = 400, a fixed positive number, and the bottom u − 20 tends to 0 through positive values.
2. A fixed positive number divided by a smaller and smaller positive number grows without bound: at u = 20.1 the image is 4020.1 = 4020 cm, and at u = 20.001 it is 400.020.001 = 400020 cm.
3. (a) So the image distance passes every length as u → 20⁺. From below, the bottom tends to 0 through negative values, so v falls without bound: at u = 19.9 it is 398−0.1 = −3980 cm. The discontinuity at u = 20 is an infinite discontinuity. Continuity at a point needs the limit there to be a number equal to the value; this limit is not a number at all, so no value given to v at u = 20 can repair it. An object sitting at the focal point makes no image.
4. For part (b), set the image distance to 100: 20uu − 20 = 100. Multiply both sides by u − 20, which is not zero for u ≠ 20, to get 20u = 100u − 2000.
5. (b) Take 20u from both sides: 80u = 2000, so u = 25 cm. Check: 20 × 2525 − 20 = 5005 = 100 cm, and v is positive, so the image is real and beyond the lens, as the question asks.

answer(a) an infinite discontinuity: the top tends to 400 while the bottom tends to 0, so v grows without bound on each side and no value at u = 20 can be its limit; (b) u = 25 cm

techniqueTypes of Discontinuity · Continuity at a Point

A step record is constant on each band, so each one-sided limit is simply the constant on that side. Where the two constants differ the record has a jump, and the value at the join tells you which side it is continuous from.

1. Just below 7 m the lift is still in the band from 3.5 m to 7 m, where the indicator reads 1, so lim_{h → 7⁻} F(h) = 1.
2. At 7 m and just above it the lift is in the next band, where the indicator reads 2, so F(7) = 2 and lim_{h → 7⁺} F(h) = 2.
3. (a) Both one-sided limits are numbers but they differ, so this is a jump discontinuity, of size 2 − 1 = 1 floor. The value F(7) = 2 agrees with the limit from the right, so F is continuous from the right at 7 m and not from the left.
4. For part (b), every band is 3.5 m tall and the ground band shows 0, so the band showing the floor number n begins at 3.5n meters.
5. (b) The indicator first shows 4 at 3.5 × 4 = 14 m. Check: at 13.9 m the reading is still 3, and it stays at 4 until 17.5 m.

answer(a) a jump discontinuity: lim_{h → 7⁻} F(h) = 1 and lim_{h → 7⁺} F(h) = 2, a jump of 1 floor, and F is continuous from the right at 7 m; (b) 14 m

techniqueTypes of Discontinuity · Continuity at a Point

A rule in three pieces has two joins, and each one has to be tested on its own. At each join take the limit from the left with the rule on the left and the limit from the right with the rule on the right, then compare them.

1. At the first join the left rule gives the value and the limit from the left: T(10) = 20 + 8 × 10 = 100, so lim_{m → 10⁻} T(m) = 100.
2. The middle rule gives lim_{m → 10⁺} T(m) = 60 + 4 × 10 = 100. (a) The value and both one-sided limits are 100, so the program is continuous at m = 10.
3. At the second join the middle rule gives lim_{m → 25⁻} T(m) = 60 + 4 × 25 = 160, and the last rule gives lim_{m → 25⁺} T(m) = 200 − 25 = 175.
4. (b) The two one-sided limits are different numbers, so the program is not continuous at m = 25: it jumps by 175 − 160 = 15 degrees Celsius.
5. Write the last rule as c − m and ask for no jump at m = 25: c − 25 = 160, so c = 185. Replacing 200 by 185 removes the jump. Check: 185 − 25 = 160, which is the temperature the middle rule reaches.

answer(a) continuous at m = 10, where both pieces give 100 degrees Celsius; (b) not continuous at m = 25: the limit from the left is 160 and the limit from the right is 175, a jump of 15 degrees Celsius, and replacing 200 by 185 removes it

techniqueContinuity at a Point · Types of Discontinuity

The theorem has two conditions: the function must be continuous on a closed interval, and the target must lie between the values at the two ends. Meeting both gives at least one crossing; a strictly decreasing graph gives exactly one.

1. T is built from an exponential and two constants, so it is continuous on the whole closed interval from m = 0 to m = 60. That is the first condition.
2. Work out the two end values: T(0) = 20 + 200 = 220 and T(60) = 20 + 200e^−2.4 = 38.14 to two decimal places.
3. (a) 100 lies between 38.14 and 220, so the intermediate value theorem gives a time m in the interval with T(m) = 100. The term 200e^−m/25 falls as m rises, so T is strictly decreasing and passes 100 once and no more.
4. For part (b), solve 20 + 200e^−m/25 = 100. Take 20 from both sides to get 200e^−m/25 = 80, then divide both sides by 200 to get e^−m/25 = 0.4.
5. Take logarithms of both sides: −m25 = ln 0.4, so m = 25 ln 2.5. (b) m = 22.9 minutes to one decimal place. Check: T(22.9) = 100.02 degrees Celsius, which is 100 to the nearest degree.

answer(a) T is continuous on the interval, T(0) = 220 and T(60) = 38.14, and 100 lies between them, so the theorem gives a time with T(m) = 100; T is strictly decreasing, so there is exactly one such time; (b) m = 25 ln 2.5 = 22.9 minutes

techniqueThe Intermediate Value Theorem · Continuity at a Point

Neither speed on its own has to take a value the question names, so work with their difference. Two speeds are equal exactly when the difference is zero, and the theorem finds a zero for a continuous function that changes sign.

1. Form the difference: d(m) = (16 − 0.2m) − (10 + 0.005m²) = 6 − 0.2m − 0.005m². Both speeds are polynomials, so d is continuous on the closed interval from m = 0 to m = 30.
2. Work out the end values: d(0) = 6, so the first runner is the faster at the start, and d(30) = 6 − 6 − 4.5 = −4.5, so the second runner is the faster at the finish.
3. (a) d is continuous and 0 lies between −4.5 and 6, so the intermediate value theorem gives an instant m in the race with d(m) = 0, which is an instant at which the two speeds are equal.
4. For part (b), solve 6 − 0.2m − 0.005m² = 0. Multiply both sides by −200 to get m² + 40m − 1200 = 0, which factorizes as (m + 60)(m − 20) = 0.
5. (b) The root m = −60 is an hour before the start and is rejected, so the instant is m = 20 minutes. Check the two speeds there: u(20) = 16 − 4 = 12 km/h and w(20) = 10 + 0.005 × 400 = 12 km/h, so both runners were at 12 km/h.

answer(a) d is continuous with d(0) = 6 and d(30) = −4.5, so the theorem gives an m in the race with d(m) = 0; (b) at m = 20 minutes, both runners at 12 km/h

techniqueThe Intermediate Value Theorem

The theorem is a promise about continuous functions only. A bank balance changes in one instant, so it is a step function, and a step function can pass from below a value to above it without ever taking that value.

1. Write the balance as a function of the day: B(d) = 420 for d < 10, and B(d) = 1020 for d ≥ 10. Those are the only two values it takes all month.
2. At the payment, lim_{d → 10⁻} B(d) = 420 and lim_{d → 10⁺} B(d) = 1020, so B has a jump discontinuity at d = 10 of size 1020 − 420 = 600.
3. (a) The intermediate value theorem asks for a function continuous on the closed interval. B is not continuous at d = 10, so the theorem does not apply and its conclusion fails: the balance takes no value strictly between $420 and $1020, and $800 is one of those values.
4. For part (b), the balance on the tenth is the old balance plus the payment, so a payment of p dollars leaves 420 + p dollars.
5. (b) Set 420 + p = 800, so p = $380. Check: 420 + 380 = 800. The account reaches $800 by landing on it, never by crossing it.

answer(a) the balance is not continuous: it jumps from $420 to $1020 on the tenth, a jump of $600, and the theorem needs a continuous function, so every value between them is skipped; (b) $380

techniqueThe Intermediate Value Theorem · Types of Discontinuity

Two opposite points are at the same temperature exactly when the difference g is zero there. Half a turn on, g reverses its sign, so whatever value it takes at the north point it takes with the opposite sign after 180 degrees, and a continuous function that changes sign has a zero.

1. The temperature varies continuously round the road, so g is continuous as well, being the difference of two continuous functions.
2. Work out g at the ends of a half turn: T(0) = 20 + 5 + 0 = 25 and T(180) = 20 − 5 + 0 = 15, so g(0) = 25 − 15 = 10 and g(180) = T(180) − T(360) = 15 − 25 = −10.
3. (a) g is continuous and 0 lies between −10 and 10, so the intermediate value theorem gives an x between 0 and 180 with g(x) = 0, which is a point at the same temperature as the point directly opposite it.
4. For part (b), simplify g. Half a turn on, cos(x + 180)^° = −cos x^° and sin(2x + 360)^° = sin 2x^°, so the two sine terms cancel and g(x) = 5cos x^° + 5cos x^° = 10cos x^°.
5. (b) 10cos x^° = 0 with x between 0 and 180 gives x = 90. The pair is the points 90 degrees and 270 degrees round the road, and T(90) = 20 + 0 + 3sin 180^° = 20 degrees Celsius. Check the other one: T(270) = 20 + 0 + 3sin 540^° = 20 as well.

answer(a) g is continuous with g(0) = 10 and g(180) = −10, so the theorem gives an x with g(x) = 0; (b) the points 90 degrees and 270 degrees round the road, both at 20 degrees Celsius

techniqueThe Intermediate Value Theorem · Continuity at a Point