Practice midterm with full solutions (MATH 203)

a) $(p \circ q)(x) = p\left(\sqrt{x + 5}\right) = \frac{2}{\sqrt{x + 5} - 3}$. The domain is read on the construction, in two steps: $q$ needs $x + 5 \ge 0$, so $x \ge -5$; then $q(x)$ must lie in the domain of $p$, which excludes $3$, and $\sqrt{x + 5} = 3$ exactly when $x + 5 = 9$, that is $x = 4$. The domain of $p \circ q$ is $[-5, 4) \cup (4, \infty)$, and $(p \circ q)(-1) = \frac{2}{2 - 3} = -2$.

$(q \circ p)(x) = \sqrt{\frac{2}{x - 3} + 5}$. First $x \ne 3$, for $p$. Then the radicand must be nonnegative, and here the algebra decides: bring it to ONE fraction, $\frac{2}{x - 3} + 5 = \frac{2 + 5(x - 3)}{x - 3} = \frac{5x - 13}{x - 3}$, and read its sign on a chart with the two critical numbers $\frac{13}{5}$ and $3$. The quotient is positive for $x < \frac{13}{5}$ (both factors negative), zero at $\frac{13}{5}$, negative between $\frac{13}{5}$ and $3$, positive for $x > 3$. The domain is $\left(-\infty, \frac{13}{5}\right] \cup (3, \infty)$, and $(q \circ p)(5) = \sqrt{1 + 5} = \sqrt 6$. Multiplying $\frac{2}{x - 3} + 5 \ge 0$ by $x - 3$ without knowing its sign is the classic error: it gives $x \ge \frac{13}{5}$, which keeps the forbidden interval $\left(\frac{13}{5}, 3\right)$ and loses every $x < \frac{13}{5}$.

b) Rewrite first: $\frac{e^x + 5}{e^x + 1} = \frac{(e^x + 1) + 4}{e^x + 1} = 1 + \frac{4}{e^x + 1}$. Since $e^x$ is increasing, so is $e^x + 1 > 1$, and its reciprocal is decreasing: $f$ is decreasing, so it never takes the same value twice, and it is one-to-one. Range: $e^x + 1$ takes every value of $(1, \infty)$, so $\frac{4}{e^x + 1}$ takes every value of $(0, 4)$ and $f$ every value of $(1, 5)$. The figure agrees: $f \to 5$ as $x \to -\infty$ and $f \to 1$ as $x \to \infty$.

Inverse: from $y = \frac{e^x + 5}{e^x + 1}$, clear the denominator, $ye^x + y = e^x + 5$. The unknown $e^x$ now appears in two terms: collect them on one side and FACTOR it out, $e^x(y - 1) = 5 - y$, so $e^x = \frac{5 - y}{y - 1}$ and $x = \ln\frac{5 - y}{y - 1}$. Renaming, $f^{-1}(x) = \ln\frac{5 - x}{x - 1}$, whose domain is the range of $f$, $(1, 5)$: exactly where the fraction is positive, as a logarithm requires. Check: $f(0) = \frac{6}{2} = 3$ and $f^{-1}(3) = \ln 1 = 0$. Then $f^{-1}(2) = \ln\frac{3}{1} = \ln 3 \approx 1.0986$, and indeed $f(\ln 3) = \frac{3 + 5}{3 + 1} = 2$. Writing $\ln\frac{5 - x}{x - 1}$ as $\frac{\ln(5 - x)}{\ln(x - 1)}$ is a false law: the quotient law gives $\ln(5 - x) - \ln(x - 1)$.

c) $9^x = (3^2)^x = (3^x)^2$ and $3^{x + 1} = 3 \cdot 3^x$: with $u = 3^x$, the equation becomes $u^2 - 6u - 27 = 0$, a quadratic hidden behind the laws of exponents. It factors as $(u - 9)(u + 3) = 0$, so $u = 9$ or $u = -3$. Back to $x$: $3^x = 9$ gives $x = 2$, and $3^x = -3$ has no solution, since $3^x > 0$ for every $x$; that candidate is rejected. Check: $81 - 2 \cdot 27 - 27 = 0$. The trap is $2 \cdot 3^{x + 1} = 6^{x + 1}$: a law of exponents never multiplies a base by a coefficient.

a) Substitution gives $\frac{\frac{1}{2} - \frac{1}{2}}{0} = \frac{0}{0}$. First gesture, the complex fraction: bring the two small fractions to one, $\frac{1}{x - 1} - \frac{1}{2} = \frac{2 - (x - 1)}{2(x - 1)} = \frac{3 - x}{2(x - 1)}$. Second gesture, factor: $x^2 - 9 = (x - 3)(x + 3)$, and $3 - x = -(x - 3)$. Dividing by $x^2 - 9$ means multiplying the denominator by it: for $x \ne 3$, the quotient is $\frac{-(x - 3)}{2(x - 1)(x - 3)(x + 3)} = \frac{-1}{2(x - 1)(x + 3)}$, and the limit is $\frac{-1}{2 \cdot 2 \cdot 6} = -\frac{1}{24}$.

Two algebra slips cost the marks here: dropping the bracket in $2 - (x - 1)$, which gives $2 - x - 1 = 1 - x$ and leaves no factor $x - 3$ to cancel, and cancelling $3 - x$ against $x - 3$ as if they were equal, which gives $+\frac{1}{24}$.

b) Substitution gives $\frac{0}{0}$. First gesture, factor the denominator: $x^2 + 5x = x(x + 5)$. Second gesture, give the sine the denominator it needs. With $\tan 2x = \frac{\sin 2x}{\cos 2x}$: $\frac{\tan 2x}{x(x + 5)} = \frac{\sin 2x}{2x} \cdot \frac{2}{(x + 5)\cos 2x}$, the factor $2$ multiplied and divided at once. As $x \to 0$, $2x \to 0$ and $2x \ne 0$, so the first factor tends to $1$, and the second tends to $\frac{2}{5 \cdot 1}$. The limit is $\frac{2}{5}$. The answer $2$ comes from dividing by $x$ alone, as if the factor $x + 5$ were not there.

c) The domain decides the side. $\sqrt{25 - x^2}$ needs $-5 \le x \le 5$ and $\sqrt{5 - x}$ in the denominator needs $x < 5$: near $5$ the function exists only on the left. The right-hand limit has no meaning, and at this endpoint of the domain only $x \to 5^-$ can be asked. Substitution gives $\frac{0}{0}$. Factor under the root, $25 - x^2 = (5 - x)(5 + x)$, and split it: $\sqrt{(5 - x)(5 + x)} = \sqrt{5 - x}\,\sqrt{5 + x}$, legal because both factors are nonnegative for $-5 \le x < 5$. The quotient equals $\sqrt{5 + x}$ there, and the limit is $\sqrt{10} \approx 3.1623$. The rule $\sqrt{ab} = \sqrt a\,\sqrt b$ needs $a \ge 0$ and $b \ge 0$: that is why the domain was read first.

a) Factor completely before deciding anything: $x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2)$, the common factor $x$ FIRST, then the difference of squares; and $x^2 + x - 6 = (x + 3)(x - 2)$. The denominator vanishes at $2$ and at $-3$, but only a zero that the numerator does not share gives a vertical asymptote. At $x = 2$ the factor cancels: for $x \ne 2$, $f(x) = \frac{x(x + 2)}{x + 3}$, so $\lim_{x \to 2} f(x) = \frac{8}{5}$, a finite limit. The graph has a hole at $\left(2, \frac{8}{5}\right)$, the open dot of the figure, and no asymptote there.

At $x = -3$, the numerator $x(x + 2)$ tends to $(-3)(-1) = 3 > 0$ while $x + 3 \to 0$. For $x \to -3^+$, $x + 3 > 0$ and $f(x) \to +\infty$; for $x \to -3^-$, $x + 3 < 0$ and $f(x) \to -\infty$. Long division of $x^2 + 2x$ by $x + 3$ gives the quotient $x - 1$ and the remainder $3$: $f(x) = x - 1 + \frac{3}{x + 3}$ for $x \ne 2$. The fraction tends to $0$ as $x \to \pm\infty$, so $y = x - 1$ is the oblique asymptote. For $x > -3$, $\frac{3}{x + 3} > 0$: the graph lies above the asymptote, and below it for $x < -3$. Dividing $x^3 - 4x$ by $x^2 + x - 6$ directly gives the same quotient $x - 1$, with the remainder $3x - 6$: the hole changes nothing at infinity.

b) The key fact is $\sqrt{x^2} = |x|$. Factor $x^2$ out of the root, $\sqrt{4x^2 + 1} = |x|\sqrt{4 + \frac{1}{x^2}}$, then divide the numerator and the denominator by $x$. As $x \to \infty$, $|x| = x$: $g(x) = \frac{\sqrt{4 + 1/x^2} + 1}{1 - 3/x} \to \frac{2 + 1}{1} = 3$. As $x \to -\infty$, $|x| = -x$, so $\frac{|x|}{x} = -1$: $g(x) = \frac{-\sqrt{4 + 1/x^2} + 1}{1 - 3/x} \to -2 + 1 = -1$. Two horizontal asymptotes, $y = 3$ on the right and $y = -1$ on the left. Writing $\frac{\sqrt{4x^2 + 1}}{x} = \sqrt{4 + \frac{1}{x^2}}$ for negative $x$ gives $3$ at both ends and loses the second one.

The denominator vanishes only at $3$. The numerator is positive for EVERY $x$, since $\sqrt{4x^2 + 1} > \sqrt{4x^2} = 2|x| \ge -x$. So the sign of $g$ is the sign of $x - 3$: $g(x) \to -\infty$ as $x \to 3^-$ and $g(x) \to +\infty$ as $x \to 3^+$, and $x = 3$ is the vertical asymptote. The same remark shows $g < 0$ for every $x < 3$, consistent with the asymptote $y = -1$ on the left.

a) For $x < 0$, $e^x \ne 1$, so the first piece is a quotient of continuous functions with a nonzero denominator; the second piece is continuous everywhere. Only the seam $x = 0$ remains, where $f(0) = c^2 - 2c$ comes from the piece with the equality sign, and the right-hand limit is the same number. The left piece gives $\frac{0}{0}$ at $0$ and must be simplified, never plugged in. The algebra: $e^{3x} = (e^x)^3$, so the numerator is a DIFFERENCE OF CUBES in $u = e^x$, $u^3 - 1 = (u - 1)(u^2 + u + 1)$. For $x < 0$, $\frac{e^{3x} - 1}{e^x - 1} = e^{2x} + e^x + 1$, which tends to $1 + 1 + 1 = 3$ as $x \to 0^-$.

Continuity at $0$ therefore requires $c^2 - 2c = 3$, that is $c^2 - 2c - 3 = (c - 3)(c + 1) = 0$: $c = 3$ or $c = -1$. Both work, and the question asked for every constant: with $c = 3$ the right piece is $9 - 6\cos x$, with $c = -1$ it is $1 + 2\cos x$, and both equal $3$ at $0$. Simplifying $\frac{e^{3x} - 1}{e^x - 1}$ term by term, as $\frac{e^{3x}}{e^x} + \frac{-1}{-1} = e^{2x} + 1$, is not algebra: it gives the limit $2$ and the wrong equation $c^2 - 2c = 2$.

b) Let $h(x) = x\,2^x - 5$; the solutions are the zeros of $h$. $h$ is the product of a polynomial and an exponential, minus a constant, so it is continuous on $\mathbb{R}$, in particular on the CLOSED interval $[1, 2]$. The values: $h(1) = 2 - 5 = -3 < 0$ and $h(2) = 8 - 5 = 3 > 0$. Since $0$ lies strictly between $h(1)$ and $h(2)$, the Intermediate Value Theorem gives a number $c$ in $(1, 2)$ with $h(c) = 0$.

Bisection, keeping each time the half where $h$ changes sign: $h(1.5) \approx -0.7574 < 0$, keep $[1.5, 2]$; $h(1.75) \approx 0.8863 > 0$, keep $[1.5, 1.75]$; $h(1.625) \approx 0.0122 > 0$, keep $[1.5, 1.625]$, of length $\frac{1}{8}$. Each step is the theorem again, on a smaller interval. The tiny value at $1.625$ says the solution is near that end (it is about $1.6231$), but three steps prove only that it lies in $[1.5, 1.625]$.

c) For $x \le 0$: $x \le 0$ and $2^x > 0$, so $x\,2^x \le 0 < 5$, and there is no solution. For $x > 0$: $x$ and $2^x$ are both positive and increasing, so their product is increasing, since $0 < a < b$ gives $a\,2^a < b\,2^a < b\,2^b$. An increasing function takes the value $5$ at most once. With b), the equation has exactly one solution, in $[1.5, 1.625]$. The theorem proves existence only: uniqueness came from this sign argument, not from the theorem.

a) The numerator $f(x + h) - f(x)$ is a difference of two fractions: put it over ONE denominator first, that is the gesture of the question. $f(x + h) - f(x) = \frac{\left((x + h)^2 + 3\right)(x - 1) - (x^2 + 3)(x + h - 1)}{(x + h - 1)(x - 1)}$. Write $A = x^2 + 3$ for short, $(x + h)^2 + 3 = A + 2xh + h^2$ and $x + h - 1 = (x - 1) + h$. The numerator becomes $(A + 2xh + h^2)(x - 1) - A(x - 1) - Ah = (2xh + h^2)(x - 1) - Ah = h\left[(2x + h)(x - 1) - (x^2 + 3)\right]$. It is a multiple of $h$, as it must be; if it is not, an error has been made upstream.

Dividing by $h \ne 0$: $\frac{f(x + h) - f(x)}{h} = \frac{(2x + h)(x - 1) - (x^2 + 3)}{(x + h - 1)(x - 1)}$. With $x$ fixed and $h \to 0$: $f'(x) = \frac{2x(x - 1) - x^2 - 3}{(x - 1)^2} = \frac{x^2 - 2x - 3}{(x - 1)^2} = \frac{(x - 3)(x + 1)}{(x - 1)^2}$, for $x \ne 1$. The quotient rule, allowed as a check, gives $\frac{2x(x - 1) - (x^2 + 3) \cdot 1}{(x - 1)^2}$, the same thing. The frequent slip is to expand $-(x^2 + 3)(x + h - 1)$ with only its first term negated.

b) $f(2) = \frac{7}{1} = 7$ and $f'(2) = \frac{(-1)(3)}{1} = -3$: the tangent is $y = 7 - 3(x - 2)$, that is $y = -3x + 13$. Horizontal tangents: $f'(x) = 0$ where the numerator vanishes and the denominator does not, $x = 3$ or $x = -1$. The points are $(3, f(3)) = \left(3, \frac{12}{2}\right) = (3, 6)$ and $(-1, f(-1)) = \left(-1, \frac{4}{-2}\right) = (-1, -2)$.

c) (i) With $F(x) = x^{2/3}$ and $a = 8$: $F(8) = \left(\sqrt[3]{8}\right)^2 = 4$ is exactly the subtracted number and the denominator is $x - 8$, so the limit is $F'(8)$ in the $x \to a$ form. With the power rule, $F'(x) = \frac{2}{3}x^{-1/3}$ and $F'(8) = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$. Algebra agrees: with $t = x^{1/3} \to 2$, the quotient is $\frac{t^2 - 4}{t^3 - 8} = \frac{t + 2}{t^2 + 2t + 4} \to \frac{4}{12}$.

(ii) With $F = \tan$ and $a = \frac{\pi}{3}$: $\tan\frac{\pi}{3} = \sqrt 3$ is the subtracted number, so the limit is $F'\left(\frac{\pi}{3}\right) = \sec^2\frac{\pi}{3} = \frac{1}{\cos^2(\pi/3)} = \frac{1}{1/4} = 4$. No algebra opens $\tan\left(\frac{\pi}{3} + h\right)$ quickly: recognizing the derivative is the move.

d) For $m$: $m(2) = 0$ and, for $h \ne 0$, $\frac{m(2 + h) - m(2)}{h} = \frac{h\,|h|}{h} = |h|$. Both one-sided limits are $0$: $m$ is differentiable at $2$, with $m'(2) = 0$. An absolute value does not always make a corner: here the factor $x - 2$ flattens it.

For $n$: $n(x) = |x|\,|x - 2|$, and near $2$, $|x| = x$. So $\frac{n(2 + h) - n(2)}{h} = \frac{(2 + h)\,|h|}{h}$, which equals $2 + h \to 2$ for $h > 0$ and $-(2 + h) \to -2$ for $h < 0$. The one-sided derivatives are $2$ and $-2$: the graph of $n$ has a corner at $(2, 0)$, and $n$ is continuous but not differentiable there.