MATH 140 practice midterm with full solutions (McGill)

a) $(p \circ q)(x) = p(x^2 + 2) = \sqrt{6 - (x^2 + 2)} = \sqrt{4 - x^2}$. The domain is read on the two steps: $q$ accepts every real $x$, and $q(x)$ must lie in the domain of $p$, that is $x^2 + 2 \le 6$, so $x^2 \le 4$ and $x \in [-2, 2]$. Then $(q \circ p)(x) = \left(\sqrt{6 - x}\right)^2 + 2 = 8 - x$. The simplified formula accepts every $x$, but the first step does not: $p(x)$ exists only for $x \le 6$. The domain of $q \circ p$ is $(-\infty, 6]$, and reading it on $8 - x$ is the classic error.

b) Domain: $e^x - 1 > 0$ means $e^x > 1$, so $x > 0$; the domain is $(0, \infty)$. Range: as $x$ runs over $(0, \infty)$, $e^x - 1$ takes every value of $(0, \infty)$, and $\ln$ takes every real value on $(0, \infty)$, so the range is $\mathbb{R}$. One-to-one: if $f(a) = f(b)$, then applying $e^{(\cdot)}$ gives $e^a - 1 = e^b - 1$, so $e^a = e^b$ and $a = b$, since $e^x$ is one-to-one.

Inverse: from $y = \ln(e^x - 1)$, $e^y = e^x - 1$, so $e^x = 1 + e^y$ and $x = \ln(1 + e^y)$. Renaming, $f^{-1}(x) = \ln(1 + e^x)$. Its domain is the range of $f$, $\mathbb{R}$, and its range is the domain of $f$, $(0, \infty)$: indeed $1 + e^x > 1$, so $\ln(1 + e^x) > 0$. Check: $f^{-1}(0) = \ln 2$ and $f(\ln 2) = \ln(2 - 1) = 0$. On the figure, the blue curve meets the $x$-axis at $\ln 2 \approx 0.69$ and the orange one meets the $y$-axis at the same height: the two marked points are mirror images in $y = x$.

c) Domain first: the three logarithms need $x > -1$, $x > 2$ and $4x - 8 > 0$, so $x > 2$. On that domain the product law applies: $\ln\left((x + 1)(x - 2)\right) = \ln\left(4(x - 2)\right)$, and since $\ln$ is one-to-one, $(x + 1)(x - 2) = 4(x - 2)$. Do not divide by $x - 2$: factor, $(x - 2)(x + 1 - 4) = 0$, so $x = 2$ or $x = 3$. The candidate $x = 2$ is outside the domain ($\ln 0$ does not exist) and is rejected. Check $x = 3$: $\ln 4 + \ln 1 = \ln 4$. The only solution is $x = 3$.

d) Let $\theta = \arccos\left(-\frac{1}{3}\right)$. The range of $\arccos$ is $[0, \pi]$, and $\cos\theta < 0$, so $\theta$ lies in $\left(\frac{\pi}{2}, \pi\right)$, where $\sin\theta > 0$: $\sin\theta = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt 2}{3}$. Then $\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \frac{2\sqrt 2}{3} \cdot \left(-\frac{1}{3}\right) = -\frac{4\sqrt 2}{9}$ and $\cos 2\theta = 2\cos^2\theta - 1 = \frac{2}{9} - 1 = -\frac{7}{9}$. So $\tan 2\theta = \frac{-4\sqrt 2/9}{-7/9} = \frac{4\sqrt 2}{7}$. The sign is consistent: $2\theta$ lies in $(\pi, 2\pi)$ with sine and cosine both negative, the third quadrant, where the tangent is positive. The formula $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$ with $\tan\theta = -2\sqrt 2$ gives $\frac{-4\sqrt 2}{-7}$, the same value.

a) Substitution gives $\frac{0}{0}$: both polynomials vanish at $3$, so both are divisible by $x - 3$. Factor: $x^2 - 2x - 3 = (x - 3)(x + 1)$ and, by the difference of cubes, $x^3 - 27 = (x - 3)(x^2 + 3x + 9)$. For $x \ne 3$ the quotient equals $\frac{x + 1}{x^2 + 3x + 9}$, a rational function defined at $3$, so the limit is $\frac{4}{9 + 9 + 9} = \frac{4}{27}$.

b) Form $\frac{0}{0}$ with a root on top: multiply top and bottom by the conjugate $\sqrt{2x + 3} + 3$. The numerator becomes $(2x + 3) - 9 = 2(x - 3)$, and $x^2 - 9 = (x - 3)(x + 3)$. For $x \ne 3$ the quotient is $\frac{2}{(x + 3)\left(\sqrt{2x + 3} + 3\right)}$, and the limit is $\frac{2}{6 \cdot 6} = \frac{1}{18}$.

c) The box of the standard limit must be the argument of the sine, $u = x^2 + 3x$, which tends to $0$ and is nonzero for $0 < |x| < 3$. Write $\frac{\sin(x^2 + 3x)}{x} = \frac{\sin u}{u} \cdot \frac{x^2 + 3x}{x} = \frac{\sin u}{u} \cdot (x + 3)$. As $x \to 0$, $u \to 0$ so the first factor tends to $1$, and the second tends to $3$: the limit is $3$. Replacing $\sin(x^2 + 3x)$ by $\sin(3x)$ because $x^2$ is small happens to give the same number here, but it is not a move you can justify.

d) The factor $\sqrt x$ exists only for $x \ge 0$, so only $x \to 0^+$ makes sense. The factor $e^{\sin(1/x)}$ has NO limit at $0$: $\sin(1/x)$ oscillates between $-1$ and $1$ infinitely often. The product law needs the limit of each factor, so it cannot be used. Squeeze: $-1 \le \sin(1/x) \le 1$, and $e^t$ is increasing, so $e^{-1} \le e^{\sin(1/x)} \le e$. Multiplying by $\sqrt x \ge 0$ keeps the inequalities: $e^{-1}\sqrt x \le \sqrt x\, e^{\sin(1/x)} \le e\sqrt x$. Both bounds tend to $0$ as $x \to 0^+$, so the limit is $0$.

a) On each open interval the formula is continuous: $\frac{\sin 6x}{2x}$ is a quotient of continuous functions whose denominator is not $0$ for $x < 0$; $ax + b$ is a polynomial; for $x > 2$, $\sqrt{x + 2} > 2$, so the last denominator is not $0$. Only the seams $0$ and $2$ remain. At $0$: $f(0) = b$, the right limit is $b$, and the left limit is $\lim_{x \to 0^-} 3 \cdot \frac{\sin 6x}{6x} = 3$, the box $6x$ matched on top and bottom. So $b = 3$.

At $2$: $f(2) = 2a + 3$ is also the left limit. The right piece is of the form $\frac{0}{0}$ at $2$ and must be simplified, never plugged in: multiplying by the conjugate $\sqrt{x + 2} + 2$, the denominator becomes $(x + 2) - 4 = x - 2$, so for $x > 2$, $\frac{x^2 - 4}{\sqrt{x + 2} - 2} = \frac{(x - 2)(x + 2)\left(\sqrt{x + 2} + 2\right)}{x - 2} = (x + 2)\left(\sqrt{x + 2} + 2\right)$, which tends to $4 \cdot 4 = 16$. Continuity at $2$ requires $2a + 3 = 16$, so $a = \frac{13}{2}$.

b) Let $h(x) = \cos x - x^2 + 1$; the solutions of the equation are the zeros of $h$. The function $h$ is the sum of $\cos$ and a polynomial, so it is continuous on $\mathbb{R}$, in particular on the closed interval $\left[0, \frac{\pi}{2}\right]$. The values: $h(0) = 1 - 0 + 1 = 2 > 0$ and $h\left(\frac{\pi}{2}\right) = 0 - \frac{\pi^2}{4} + 1 < 0$, because $\pi > 2$ gives $\pi^2 > 4$. Since $0$ lies strictly between $h\left(\frac{\pi}{2}\right)$ and $h(0)$, the Intermediate Value Theorem gives $c$ in $\left(0, \frac{\pi}{2}\right)$ with $h(c) = 0$.

Second solution: $h$ is even, since $\cos(-x) = \cos x$ and $(-x)^2 = x^2$. So $h(-c) = h(c) = 0$, and $-c \ne c$ because $c > 0$. (Equally, the theorem applies on $\left[-\frac{\pi}{2}, 0\right]$.) The figure agrees: the two curves cross twice, symmetrically, a little to each side of $x = \pm 1$.

c) $h\left(-\frac{\pi}{2}\right) = h\left(\frac{\pi}{2}\right) = 1 - \frac{\pi^2}{4} < 0$: same sign at both ends, so the theorem gives NOTHING on that interval. It does not say that there is no zero; it is silent, and here there are two zeros inside, one on each side of $0$. Location: if $\cos x = x^2 - 1$, then $x^2 - 1 \le 1$ because $\cos x \le 1$, so $x^2 \le 2$ and $|x| \le \sqrt 2$. The theorem is an existence theorem: it guarantees at least one zero on each of the two intervals, and says nothing about how many there are. Proving that there are exactly two needs a tool from later in the course.

a) Factor before deciding anything: $2x^2 + 3x - 5 = (2x + 5)(x - 1)$ and $x^2 - 1 = (x - 1)(x + 1)$. The denominator vanishes at $1$ and at $-1$, but only a zero that the numerator does NOT share can give a vertical asymptote. At $x = 1$ the factor cancels: for $x \ne \pm 1$, $f(x) = \frac{2x + 5}{x + 1}$, so $\lim_{x \to 1} f(x) = \frac{7}{2}$, a finite limit. The graph has a hole at $\left(1, \frac{7}{2}\right)$, the open dot of the figure, and no asymptote there.

At $x = -1$ the numerator $2x + 5$ tends to $3 > 0$ and $x + 1 \to 0$. For $x \to -1^+$, $x + 1 > 0$, so $f(x) \to +\infty$; for $x \to -1^-$, $x + 1 < 0$, so $f(x) \to -\infty$. The line $x = -1$ is a vertical asymptote. At infinity, divide by $x^2$: $f(x) = \frac{2 + \frac{3}{x} - \frac{5}{x^2}}{1 - \frac{1}{x^2}} \to 2$ as $x \to \pm\infty$, so $y = 2$ is the only horizontal asymptote. Writing $f(x) = 2 + \frac{3}{x + 1}$ shows the curve above it for $x > -1$ and below it for $x < -1$, as drawn.

b) The key fact is $\sqrt{x^2} = |x|$. For $x \ne 0$ in the domain, $\sqrt{9x^2 + x} = \sqrt{x^2\left(9 + \frac{1}{x}\right)} = |x|\sqrt{9 + \frac{1}{x}}$. As $x \to \infty$, $|x| = x$ and $g(x) = \frac{x\sqrt{9 + 1/x}}{x\left(1 + 2/x\right)} = \frac{\sqrt{9 + 1/x}}{1 + 2/x} \to 3$. As $x \to -\infty$, $|x| = -x$, and $g(x) = \frac{-\sqrt{9 + 1/x}}{1 + 2/x} \to -3$. Sign check: for large negative $x$ the numerator is positive and $x + 2$ is negative, so $g < 0$. Two horizontal asymptotes: $y = 3$ on the right and $y = -3$ on the left. Writing $\frac{\sqrt{9x^2 + x}}{x} = \sqrt{9 + \frac{1}{x}}$ for negative $x$ is the error that loses the second one.

c) A form $\infty - \infty$, which is not a number: multiply and divide by the conjugate. $\sqrt{x^2 + 7x} - \sqrt{x^2 + x} = \frac{(x^2 + 7x) - (x^2 + x)}{\sqrt{x^2 + 7x} + \sqrt{x^2 + x}} = \frac{6x}{\sqrt{x^2 + 7x} + \sqrt{x^2 + x}}$. For $x > 0$, divide top and bottom by $x = \sqrt{x^2}$: $\frac{6}{\sqrt{1 + \frac{7}{x}} + \sqrt{1 + \frac{1}{x}}} \to \frac{6}{1 + 1} = 3$. The answer $0$, from $\sqrt{x^2 + 7x} \approx x \approx \sqrt{x^2 + x}$, forgets that the two errors are each of the size of a constant.

a) $f(1) = \frac{1}{2}$, so the quotient is $\frac{1}{h}\left(\frac{1}{\sqrt{4 + h}} - \frac{1}{2}\right) = \frac{2 - \sqrt{4 + h}}{2h\sqrt{4 + h}}$ after a common denominator. It is still $\frac{0}{0}$, with a root on top: multiply by the conjugate $2 + \sqrt{4 + h}$. The numerator becomes $4 - (4 + h) = -h$, and for $h \ne 0$ the quotient is $\frac{-1}{2\sqrt{4 + h}\left(2 + \sqrt{4 + h}\right)}$. Letting $h \to 0$: $f'(1) = \frac{-1}{2 \cdot 2 \cdot 4} = -\frac{1}{16}$.

Tangent line: $y = \frac{1}{2} - \frac{1}{16}(x - 1)$, that is $y = -\frac{x}{16} + \frac{9}{16}$. Check with the power rule, allowed as a check only: $f(x) = (x + 3)^{-1/2}$ gives $-\frac{1}{2}(x + 3)^{-3/2}$, which is $-\frac{1}{2} \cdot \frac{1}{8} = -\frac{1}{16}$ at $x = 1$.

b) $g(x + h) - g(x) = \frac{1}{(x + h)^2 + 1} - \frac{1}{x^2 + 1} = \frac{x^2 + 1 - (x + h)^2 - 1}{\left((x + h)^2 + 1\right)(x^2 + 1)} = \frac{-2xh - h^2}{\left((x + h)^2 + 1\right)(x^2 + 1)}$. Dividing by $h \ne 0$, the factor $h$ cancels: $\frac{g(x + h) - g(x)}{h} = \frac{-2x - h}{\left((x + h)^2 + 1\right)(x^2 + 1)}$. With $x$ fixed and $h \to 0$, $g'(x) = \frac{-2x}{(x^2 + 1)^2}$. The denominator never vanishes, so $g$ is differentiable at every real $x$. Sanity check: $g$ is even and peaks at $x = 0$, and indeed $g'(0) = 0$ and $g'$ is odd.

c) (i) With $F(x) = x^{10}$ and $a = 1$, $F(1) = 1$ is the subtracted constant, so the limit is the second form of $F'(1)$, and $F'(1) = 10 \cdot 1^9 = 10$. Algebra agrees: $x^{10} - 1 = (x - 1)(x^9 + x^8 + \dots + x + 1)$, and the ten terms of the second factor each tend to $1$.

(ii) With $F = \arctan$ and $a = 1$, $F(1) = \arctan 1 = \frac{\pi}{4}$ is the subtracted constant: the limit is $F'(1)$ in the $h$ form. Since $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$, it equals $\frac{1}{2}$. No algebra could open $\arctan(1 + h)$; recognizing the derivative is the only way.

d) Continuity at $1$ comes first, because a function that is differentiable at a point is continuous there. The left value is $k(1) = 1$ and the right limit is $a + b$, so $a + b = 1$. If this fails, the right quotient $\frac{a(1 + h) + b - 1}{h} = \frac{(a + b - 1) + ah}{h}$ tends to $\pm\infty$ and no slope exists. With $a + b = 1$, the right quotient is exactly $\frac{ah}{h} = a$, and the left quotient is $\frac{(1 + h)^2 - 1}{h} = 2 + h \to 2$. The two one-sided derivatives are equal when $a = 2$, so $a = 2$ and $b = -1$: the line $y = 2x - 1$ is the tangent to the parabola at $(1, 1)$, and the two pieces join smoothly.

With $a = 3$ and $b = -2$, still $a + b = 1$: $k$ is continuous at $1$, but the one-sided slopes are $2$ and $3$. The graph has a corner at $(1, 1)$ and $k$ is not differentiable there. Continuity is necessary, not sufficient.

a) Expand and divide term by term, with $x\sqrt x = x^{3/2}$: $y = \frac{x^2 + 2x + 1}{x^{3/2}} = x^{1/2} + 2x^{-1/2} + x^{-3/2}$. Power rule, term by term: $y' = \frac{1}{2}x^{-1/2} - x^{-3/2} - \frac{3}{2}x^{-5/2}$. Over the common denominator $2x^{5/2}$: $y' = \frac{x^2 - 2x - 3}{2x^{5/2}} = \frac{(x - 3)(x + 1)}{2x^{5/2}}$. Then $y'(1) = \frac{(-2)(2)}{2} = -2$. For $x > 0$ the factors $x + 1$ and $x^{5/2}$ are positive, so $y' = 0$ only at $x = 3$, where $y = \frac{16}{3\sqrt 3} = \frac{16\sqrt 3}{9}$.