Corrected exercises: rates of change and limit laws (MATH 203)

a) Substituting: $f(-1) = -1 - 3 + 5 = 1$, $f(0) = 5$, $f(2) = 8 - 12 + 5 = 1$, $f(3) = 27 - 27 + 5 = 5$, which the grid confirms. On $[-1, 0]$: $\frac{5 - 1}{0 - (-1)} = 4$. On $[0, 2]$: $\frac{f(2) - f(0)}{2 - 0} = \frac{1 - 5}{2} = -2$. On $[-1, 3]$: $\frac{5 - 1}{3 - (-1)} = \frac{4}{4} = 1$, and this is the slope of the dashed line, which joins $(-1, 1)$ to $(3, 5)$. Subtracting a negative abscissa is the first algebraic trap: $3 - (-1) = 4$, not $2$. The two classic slips are the inverted quotient $\frac{\Delta x}{\Delta y}$, which would give $\frac{1}{4}$ on $[-1, 0]$, and the mixed order $\frac{f(0) - f(2)}{2 - 0} = 2$, which loses the sign. A negative rate on $[0, 2]$ is exactly what the figure shows: the curve comes DOWN from $5$ to $1$.

b) The rate on $[0, b]$ is $\frac{f(b) - 5}{b - 0}$, which is $0$ exactly when $f(b) = 5$, that is $b^3 - 3b^2 = 0$. Factor out the common factor: $b^2(b - 3) = 0$, so $b = 0$ or $b = 3$, and the only root in $(0, 3.2]$ is $b = 3$. Dividing both sides by $b^2$ instead of factoring gives the right root here by luck, and loses roots in general: an equation is FACTORED, never divided by an expression that may be $0$. A zero average rate says that $f$ ends where it started: the secant is HORIZONTAL. It does not say that $f$ stayed constant, nor that it did not move: on $[0, 3]$ the curve goes down to $1$ and back up to $5$. The same holds on $[-1, 2]$, since $f(-1) = f(2) = 1$.

c) With $x$ in radians: on $\left[0, \frac{\pi}{6}\right]$, $\frac{\sin\frac{\pi}{6} - \sin 0}{\frac{\pi}{6} - 0} = \frac{\frac{1}{2}}{\frac{\pi}{6}} = \frac{1}{2} \cdot \frac{6}{\pi} = \frac{3}{\pi} \approx 0.9549$. On $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$: $\frac{1 - \frac{1}{2}}{\frac{\pi}{2} - \frac{\pi}{6}} = \frac{\frac{1}{2}}{\frac{\pi}{3}} = \frac{3}{2\pi} \approx 0.4775$. The second is half the first: the sine curve flattens as it approaches its maximum. Dividing by a fraction is multiplying by its reciprocal; this is where the marks go, not in the sine. A calculator in degree mode reads the interval as $[0, 30]$ and returns $\frac{0.5}{30} \approx 0.0167$, which is the rate per DEGREE. In calculus every angle is in radians unless the question says otherwise.

d) On $[2, 5]$: $\frac{\frac{1}{5} - \frac{1}{2}}{5 - 2} = \frac{-\frac{3}{10}}{3} = -\frac{1}{10}$. In general, the numerator is a difference of fractions: bring it to the common denominator $ab$ first, $\frac{1}{b} - \frac{1}{a} = \frac{a - b}{ab}$. Then divide by $b - a$, that is multiply by $\frac{1}{b - a}$: $\frac{a - b}{ab(b - a)}$. Since $a - b = -(b - a)$, the factor cancels and leaves $-\frac{1}{ab}$. Check with $a = 2$, $b = 5$: $-\frac{1}{10}$. This fraction inside a fraction is THE algebraic gesture of the chapter; writing $\frac{1}{b} - \frac{1}{a} = \frac{1}{b - a}$ is the error that costs the whole part.

e) $t^{-2} = \frac{1}{t^2}$, never $-t^2$. On $[1, 2]$: $\frac{\frac{1}{4} - 1}{2 - 1} = -\frac{3}{4}$. On $[1, 1 + k]$: $\frac{(1 + k)^{-2} - 1}{k} = \frac{\frac{1}{(1 + k)^2} - 1}{k} = \frac{1 - (1 + k)^2}{k(1 + k)^2}$. Expand the top: $1 - (1 + 2k + k^2) = -2k - k^2 = -k(2 + k)$. The factor $k$ cancels ($k \ne 0$): the rate is $-\frac{2 + k}{(1 + k)^2}$. For $k = 0.1$: $-\frac{2.1}{1.21} \approx -1.7355$. Two traps: $(1 + k)^{-2}$ is not $1 + k^{-2}$, and $(1 + k)^2$ is not $1 + k^2$; with the second one, the $k$ never factors out and the simplification stalls.

a) The slope of $PQ$ with $Q(1 + h, 2^{1 + h})$ is $\frac{2^{1 + h} - 2}{h}$. With $h = 1$: $\frac{4 - 2}{1} = 2$. With $h = 0.5$: $\frac{2^{1.5} - 2}{0.5} = \frac{2.828427 - 2}{0.5} \approx 1.6569$. With $h = 0.1$: $\frac{2.143547 - 2}{0.1} \approx 1.4355$. With $h = 0.01$: $\frac{2.013911 - 2}{0.01} \approx 1.3911$. With $h = 0.001$: $\approx 1.3868$. Keep all the digits of $2^{1 + h}$ in the calculator until the last division: rounding $2^{1.001}$ to $2.001$ before subtracting destroys the estimate, since the difference with $2$ is only $0.0014$. The figure checks the first two: $PQ_1$ is steeper than $PQ_2$.

b) Now $h$ is negative, and the formula is unchanged. $h = -0.1$: $\frac{2^{0.9} - 2}{-0.1} = \frac{1.866066 - 2}{-0.1} \approx 1.3393$. $h = -0.01$: $\approx 1.3815$. $h = -0.001$: $\approx 1.3858$. A negative numerator over a negative denominator gives a positive slope, as the rising curve demands. $Q$ cannot be $P$ itself: a secant needs two distinct points, and with $h = 0$ the quotient is $\frac{0}{0}$, which is not a number. This is precisely why the slope of a tangent is a LIMIT, the value the secant slopes approach as $h \to 0$ with $h \ne 0$.

c) The right-hand slopes decrease, $2$, $1.6569$, $1.4355$, $1.3911$, $1.3868$; the left-hand slopes increase, $1.3393$, $1.3815$, $1.3858$. Both lists close in on the same number, trapped between $1.3858$ and $1.3868$, so $m \approx 1.386$, that is $1.39$ to $2$ decimals. The curve $y = 2^x$ bends upward: a chord to the right of $P$ climbs more steeply than the tangent, a chord to the left climbs less steeply. That is why one side overestimates and the other underestimates, and why an estimate from ONE side only, such as $1.4355$ from $h = 0.1$, is off by $0.05$ without warning. With both sides you get a bracket, not just a guess.

d) Point-slope form through $P(1, 2)$: $y - 2 = 1.386(x - 1)$, so $y = 1.386x + 0.614$. The $y$-intercept is $2 - 1.386 \approx 0.61$. With the rounded slope $1.39$ one gets $0.61$ as well. A student who forgets the parentheses and writes $y = 2 + 1.386x - 1$ distributes nothing and finds the intercept $1$: the point-slope form must be expanded as $1.386x - 1.386$.

e) By the law of exponents $a^{m + n} = a^m a^n$: $2^{1 + h} = 2 \cdot 2^h$, so $2^{1 + h} - 2 = 2 \cdot 2^h - 2 = 2(2^h - 1)$, after factoring out the common $2$. Dividing by $h$ gives the identity. The right side contains $\frac{2^h - 1}{h}$, which is the slope of the secant from $(0, 1)$ to $(h, 2^h)$. So every secant slope at $P$ is exactly twice the corresponding secant slope at $(0, 1)$, and their limits keep that ratio: the slope at $(0, 1)$ is about $\frac{1.386}{2} = 0.693$, that is $0.69$. The classic slip is $2^{1 + h} = 2 + 2^h$, which turns a product into a sum and makes the factorization impossible.

a) As $x$ approaches $1$, the blue line runs toward the empty dot at height $2$: $\lim_{x\to 1} f(x) = 2$. The value is the full dot: $f(1) = 4$. The limit reads the curve AROUND $1$, the value reads the single isolated dot; they are two different questions. At $-1$ nothing special happens: $f(x) = \frac{x + 3}{2}$ near $-1$, so the limit is $\frac{-1 + 3}{2} = 1$. Near $3$, the values of $f$ just below $3$ are close to $3$ (the line ends at the full dot $(3, 3)$), while the values just above $3$ all equal $2$. The values of $f$ do not approach ONE number, so $\lim_{x\to 3} f(x)$ does not exist.

b) From the figure, $\lim_{x\to 1} f(x) = 2$ and $\lim_{x\to 1} g(x) = g(1) = 0$, since $g$ is a polynomial; its double root makes the orange curve touch the axis at $(1, 0)$. Both limits exist, so the difference and constant multiple laws apply: $3 \cdot 2 - 2 \cdot 0 = 6$. The power law gives $\lim_{x\to 1} [f(x)]^3 = 2^3 = 8$. The value $[f(1)]^3 = 4^3 = 64$ has nothing to do with it: the power law takes the power of the LIMIT, and the limit ignores the full dot at $(1, 4)$.

c) At $-1$: $\lim f(x) = 1$ and $\lim g(x) = g(-1) = \frac{(2)^2(10)}{40} = 1 \ne 0$, so the quotient law has its permit and gives $\frac{1}{1} = 1$; on the figure the two graphs cross at $(-1, 1)$. At $1$: $\lim g(x) = 0$ and $\lim f(x) = 2 \ne 0$, so the quotient $\frac{g}{f}$ tends to $\frac{0}{2} = 0$. For $\frac{f}{g}$ at $1$ the denominator tends to $0$: the quotient law has NO permit, and cannot be used. What this quotient does near $1$ (it grows without bound, since $g$ is tiny and positive near $1$) belongs to the chapter on infinite limits; here the only correct statement is that the law does not apply.

d) $g$ is a polynomial, so $\lim_{x\to 3} g(x) = g(3) = \frac{(-2)^2(6)}{40} = \frac{3}{5}$. The product law cannot be used for $fg$, since $f$ has no limit at $3$. Suppose $\lim_{x\to 3} f(x)g(x) = L$. Near $3$, $g(x)$ is close to $\frac{3}{5}$, hence not $0$, and $f(x) = \frac{f(x)g(x)}{g(x)}$. Both pieces now have limits, $L$ and $\frac{3}{5} \ne 0$, so the quotient law WOULD give $\lim_{x\to 3} f(x) = \frac{5L}{3}$, a number, contradicting a). So $\lim_{x\to 3} f(x)g(x)$ does not exist. The law used backwards is a proof technique: a factor with a nonzero limit cannot rescue a factor that has none.

e) Denominator first, since it decides the permit: $\lim (f - g + 1) = 2 - 0 + 1 = 3 \ne 0$ by the sum and difference laws. Numerator: $\lim ([f]^2 + g) = 2^2 + 0 = 4$ by the power and sum laws. Quotient law: $\frac{4}{3}$. With the values at $1$ one would compute $\frac{4^2 + 0}{4 - 0 + 1} = \frac{16}{5}$: this is $F(1)$ for the function $F$ of the question, not its limit, because $f(1) = 4$ is not the height the blue curve approaches. Substituting is legitimate only when the function IS its limit at the point, as for polynomials; $f$ is not one of them at $x = 1$.

a) Substitution: $\frac{16 - 8 - 8}{16 - 16} = \frac{0}{0}$, so $x - 4$ divides both. $x^2 - 2x - 8 = (x - 4)(x + 2)$, two numbers with product $-8$ and sum $-2$, and $x^2 - 16 = (x - 4)(x + 4)$, a difference of squares. For $x \ne 4$ the quotient is $\frac{x + 2}{x + 4}$, a rational function whose denominator is $8 \ne 0$ at $4$, so the limit is $\frac{6}{8} = \frac{3}{4}$. Reduce the fraction at the end; $\frac{6}{8}$ is correct but a marker wants the reduced form.

b) Substitution: $\frac{-1 + 1}{1 - 3 + 2} = \frac{0}{0}$. The numerator is a SUM of cubes, $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$: $x^3 + 1 = (x + 1)(x^2 - x + 1)$. The denominator is $(x + 1)(x + 2)$. For $x \ne -1$ the quotient is $\frac{x^2 - x + 1}{x + 2}$, which gives $\frac{1 + 1 + 1}{1} = 3$. Writing $x^3 + 1 = (x + 1)^3$ is false: at $x = 1$ the left side is $2$ and the right side $8$. Test a factorization at one simple value before cancelling; it takes five seconds.

c) Substitution: $27 - 18 - 12 + 3 = 0$ on top and $0$ below, the form $\frac{0}{0}$. By the factor theorem, $x - 3$ divides the cubic. Synthetic division by $3$ on the coefficients $1, -2, -4, 3$: bring down $1$; $1 \cdot 3 - 2 = 1$; $1 \cdot 3 - 4 = -1$; $-1 \cdot 3 + 3 = 0$, the remainder, as expected. So $x^3 - 2x^2 - 4x + 3 = (x - 3)(x^2 + x - 1)$, and for $x \ne 3$ the quotient is $x^2 + x - 1$, a polynomial: the limit is $9 + 3 - 1 = 11$. A remainder that is not $0$ means an arithmetic slip in the division, never a quotient that has no limit.

d) Substitution: $8 - 8 + 6 - 6 = 0$ over $4 - 4 = 0$. Group the numerator in pairs: $x^3 - 2x^2 + 3x - 6 = x^2(x - 2) + 3(x - 2) = (x - 2)(x^2 + 3)$. The common factor $x - 2$ appears only because the two pairs were chosen so that their brackets match. The denominator is $(x - 2)(x + 2)$. For $x \ne 2$ the quotient is $\frac{x^2 + 3}{x + 2}$, which gives $\frac{7}{4}$. Grouping is the fastest route when the coefficients come in proportional pairs, here $1 : -2$ and $3 : -6$.

e) Substitution: $18 - 15 - 3 = 0$ over $9 - 9 = 0$. The numerator vanishes at $-3$, so $x + 3$ is a factor, and matching the leading coefficient $2$ gives $2x^2 + 5x - 3 = (2x - 1)(x + 3)$; expanding checks it: $2x^2 + 6x - x - 3$. The denominator is $x(x + 3)$. For $x \ne -3$, the quotient is $\frac{2x - 1}{x}$, which gives $\frac{-7}{-3} = \frac{7}{3}$. The student's product $(x + 3)\left(x - \frac{1}{2}\right) = x^2 + \frac{5}{2}x - \frac{3}{2}$ is HALF the numerator: the roots are right, but a quadratic with leading coefficient $a$ factors as $a(x - r_1)(x - r_2)$, and the lost $2$ halves the answer. Expanding before cancelling catches it.