Corrected exercises: inverse functions and logarithms (MATH 203)

The line $y = 2$ meets the curve at the two marked points $(-1, 2)$ and $(2, 2)$. Check: $p(-1) = -1 + 3 = 2$ and $p(2) = 8 - 6 = 2$. Two different inputs give the same output, so $p$ is NOT one-to-one on $\mathbb{R}$, and it has no inverse there. For $p(x) = 1$, draw the line $y = 1$ in your head: it lies between the low turning value $-2$ and the high one $2$, so it cuts the three pieces of the curve, and the equation has $3$ solutions. The values $p(-2) = -2$, $p(-1) = 2$, $p(1) = -2$, $p(2) = 2$ alternate around $1$, which confirms one crossing in each of the three intervals. One counterexample settles a no; the figure is enough to find it, the computation is what proves it.

To the right of the turning point $(1, -2)$ the curve only rises, so $[1, \infty)$ is the largest interval containing $3$ on which $p$ is one-to-one; between the two turning points it only falls, so for $0$ the answer is $[-1, 1]$, where $p$ takes each value of $[-2, 2]$ once. The algebraic proof uses the gesture of the whole chapter, FACTOR before judging a sign: $p(b) - p(a) = (b^3 - a^3) - 3(b - a) = (b - a)(a^2 + ab + b^2) - 3(b - a) = (b - a)(a^2 + ab + b^2 - 3)$. For $1 \le a < b$: $a^2 \ge 1$, $ab > a^2 \ge 1$ and $b^2 > 1$, so the second factor is positive, and so is $b - a$. Hence $p(b) > p(a)$: $p$ is increasing on $[1, \infty)$, and an increasing function never repeats a value. Without the common factor $b - a$ pulled out, the sign of $b^3 - a^3 - 3b + 3a$ cannot be read, and this is where the marks go.

$q_1$: if $1 + e^{-a} = 1 + e^{-b}$ then $e^{-a} = e^{-b}$, and since the exponential is one-to-one, $-a = -b$, so $a = b$: one-to-one. $q_2(-1) = q_2(1) = 0$: not one-to-one, an even function never is on a domain symmetric about $0$. $q_3(x) = x|x|$ equals $x^2$ for $x \ge 0$ and $-x^2$ for $x < 0$, so its sign is the sign of $x$. If $a|a| = b|b|$, then $a$ and $b$ have the same sign and $a^2 = |a|\,|a| = |b|\,|b| = b^2$, so $|a| = |b|$, and with the same sign, $a = b$: one-to-one, although it is built from $x^2$. $q_4(-4) = q_4(4) = 0$: the upper half of a circle of radius $4$ fails the horizontal line test.

Read the table backwards: $f^{-1}(4)$ is the input whose output is $4$, so $f^{-1}(4) = 1$; likewise $f^{-1}(-7) = 6$ and $f^{-1}(5) = 0$. Next, $f^{-1}(f(3)) = f^{-1}(-1) = 3$, as the cancellation law $f^{-1}(f(x)) = x$ predicts on the domain of $f$. Finally $\frac{1}{f(0)} = \frac{1}{5}$, which has nothing to do with $f^{-1}(5) = 0$: the $-1$ in $f^{-1}$ means the inverse for composition, never the reciprocal. The reciprocal of the value is written $(f(0))^{-1}$, with the brackets.

(i) TRUE: if $f$ is increasing and $a < b$, then $f(a) < f(b)$, so two different inputs never share an output. (ii) FALSE: $g(x) = \frac{1}{x}$ is one-to-one, since $\frac{1}{a} = \frac{1}{b}$ gives $a = b$, but it is not decreasing on its whole domain: $-1 < 1$ and $g(-1) = -1 < 1 = g(1)$. It decreases on each of its two pieces, which is not the same thing. The one-to-one property is a statement about VALUES; a monotonicity argument is only one way to prove it.

Write $y = \frac{1 - 3x}{x + 4}$ and clear the denominator: $y(x + 4) = 1 - 3x$, that is $xy + 4y = 1 - 3x$. Now the gesture: the unknown $x$ is in two terms, $xy$ and $-3x$. Collect them on the left, everything else on the right: $xy + 3x = 1 - 4y$. Factor: $x(y + 3) = 1 - 4y$, and divide, which requires $y \ne -3$: $x = \frac{1 - 4y}{y + 3}$. Renaming, $f^{-1}(x) = \frac{1 - 4x}{x + 3}$. Its domain is $x \ne -3$, and that is the range of $f$: the value $-3$ is never taken, since $\frac{1 - 3x}{x + 4} = -3$ would give $1 - 3x = -3x - 12$, that is $1 = -12$. Its range is $y \ne -4$, the domain of $f$. Then $f^{-1}(1) = \frac{-3}{4}$, and the check: $f\left(-\frac{3}{4}\right) = \frac{1 + \frac{9}{4}}{\frac{13}{4}} = \frac{13/4}{13/4} = 1$. The classic wrong line is $x = \frac{1 - 3x - 4y}{y}$: $x$ is still on both sides, and nothing has been solved.

$g$ is decreasing, so it is one-to-one. Its range: $\sqrt{x - 1}$ takes every value in $[0, \infty)$, so $g$ takes every value in $(-\infty, 2]$. Solve $y = 2 - \sqrt{x - 1}$: $\sqrt{x - 1} = 2 - y$. A square root is never negative, so this step REQUIRES $2 - y \ge 0$, that is $y \le 2$. Squaring, $x - 1 = (2 - y)^2$, so $g^{-1}(x) = 1 + (2 - x)^2$ for $x \le 2$. The condition is part of the formula: $1 + (2 - x)^2$ on all of $\mathbb{R}$ is a full parabola, which fails the horizontal line test and is the inverse of nothing. Then $g^{-1}(0) = 1 + 4 = 5$, checked by $g(5) = 2 - 2 = 0$, and $g^{-1}(-1) = 1 + 9 = 10$, checked by $g(10) = 2 - 3 = -1$.

The domain of $h$ is $x > -3$. Isolate the logarithm first, then undo it: $y = 1 + 2\ln(x + 3)$ gives $\ln(x + 3) = \frac{y - 1}{2}$, so $x + 3 = e^{(y - 1)/2}$ and $x = e^{(y - 1)/2} - 3$. So $h^{-1}(x) = e^{(x - 1)/2} - 3$, defined for every real $x$ (the range of $h$ is $\mathbb{R}$, like that of $\ln$), with range $(-3, \infty)$, the domain of $h$, because $e^{(x-1)/2} > 0$. Values: $h^{-1}(1) = e^0 - 3 = -2$, checked by $h(-2) = 1 + 2\ln 1 = 1$; and $h^{-1}(5) = e^2 - 3 \approx 4.3891$. The frequent slip is to exponentiate before isolating: $e^y = e \cdot 2(x + 3)$ confuses $e^{2\ln(x+3)} = (x + 3)^2$ with a product.

Write $y = \frac{2^x + 1}{2^x - 1}$ and clear the denominator: $y \cdot 2^x - y = 2^x + 1$. The unknown quantity $2^x$ appears in two terms: collect them, $y \cdot 2^x - 2^x = y + 1$, factor, $2^x(y - 1) = y + 1$, and divide ($y \ne 1$): $2^x = \frac{y + 1}{y - 1}$. Now the domain appears by itself: $2^x > 0$, so we need $\frac{y + 1}{y - 1} > 0$, that is $y < -1$ or $y > 1$ (sign table of $y + 1$ and $y - 1$). Then $x = \log_2 \frac{y + 1}{y - 1}$ and $m^{-1}(x) = \log_2 \frac{x + 1}{x - 1}$ on $(-\infty, -1) \cup (1, \infty)$, which is therefore the range of $m$. Values: $m^{-1}(3) = \log_2 2 = 1$, checked by $m(1) = \frac{3}{1} = 3$; $m^{-1}(-3) = \log_2 \frac{-2}{-4} = \log_2 \frac{1}{2} = -1$, checked by $m(-1) = \frac{3/2}{-1/2} = -3$. And $m^{-1}(0)$ does NOT exist: it would be $\log_2(-1)$, and indeed $m(x) = 0$ would need $2^x = -1$.

Substitute $f^{-1}(x) = \frac{1 - 4x}{x + 3}$ into $f$: $f(f^{-1}(x)) = \dfrac{1 - 3\cdot\frac{1 - 4x}{x + 3}}{\frac{1 - 4x}{x + 3} + 4}$. The algebra gesture for a complex fraction: multiply the top AND the bottom by the inner denominator $x + 3$. Top: $(x + 3) - 3(1 - 4x) = x + 3 - 3 + 12x = 13x$. Bottom: $(1 - 4x) + 4(x + 3) = 1 - 4x + 4x + 12 = 13$. So $f(f^{-1}(x)) = \frac{13x}{13} = x$. It is valid wherever $f^{-1}(x)$ exists, that is for every $x \ne -3$. The two usual errors are multiplying only the top by $x + 3$, and distributing the $-3$ to the first term only, $x + 3 - 3 - 4x$; both break the cancellation that proves the inverse is right.

$f^{-1}(-1)$ is the INPUT whose output is $-1$: the marked point $(1, -1)$ gives $f^{-1}(-1) = 1$. Likewise $(2, 1)$ gives $f^{-1}(1) = 2$. Solving $f(x) = -2$ is the same question once more, $x = f^{-1}(-2)$, and the point $(0, -2)$ gives $x = 0$; check, $2^0 - 3 = -2$. Reading an inverse on a graph means entering on the VERTICAL axis and leaving on the horizontal one: students who enter on the horizontal axis read $f(1) = -1$ and $f(-1) = -\frac{5}{2}$ instead.

$f$ is defined for every real $x$. Since $2^x$ takes every value in $(0, \infty)$, $2^x - 3$ takes every value in $(-3, \infty)$: that is the range of $f$. Swap them: $f^{-1}$ has domain $(-3, \infty)$ and range $\mathbb{R}$. The horizontal line $y = -3$ that $f$ approaches on the left is made of points $(x, -3)$; reflected in $y = x$ they become $(-3, x)$, the VERTICAL line $x = -3$, which the graph of $f^{-1}$ approaches from the right as it falls. Domain and range swap, and so do horizontal and vertical features.

Solve $y = 2^x - 3$: $2^x = y + 3$, and apply $\log_2$, legal since $y + 3 > 0$ on the domain: $x = \log_2(y + 3)$. So $f^{-1}(x) = \log_2(x + 3)$. Then $f^{-1}(4) = \log_2 7$, the exact answer. The calculator has no $\log_2$ key, and this is exactly what the change of base formula is for: $\log_2 7 = \frac{\ln 7}{\ln 2} \approx \frac{1.945910}{0.693147} \approx 2.8074$. Sanity check: $4 < 7 < 8$, so $\log_2 7$ must lie between $2$ and $3$; and $2^{2.8074} - 3 \approx 4$. Typing $\frac{\ln 2}{\ln 7} \approx 0.3562$ is the upside-down version, and the bracket catches it at once.

$r(-1) = r(1) = e^0 = 1$, so $r$ is not one-to-one on $\mathbb{R}$. On $(-\infty, 0]$, $1 - x^2$ increases up to $1$, so $r$ increases from values close to $0$ up to $r(0) = e$: its range there is $(0, e]$, the domain of the inverse. Solve $y = e^{1 - x^2}$: take $\ln$, $\ln y = 1 - x^2$, so $x^2 = 1 - \ln y$ and $x = \pm\sqrt{1 - \ln y}$. The RESTRICTION chooses the sign: $x \le 0$, so $x = -\sqrt{1 - \ln y}$, and $r^{-1}(x) = -\sqrt{1 - \ln x}$ for $0 < x \le e$ (the square root needs $\ln x \le 1$, the same condition). Values: $r^{-1}(1) = -\sqrt{1 - 0} = -1$ and $r^{-1}(e^{-3}) = -\sqrt{1 + 3} = -2$; check, $r(-2) = e^{1 - 4} = e^{-3}$. Answering $+1$ gives a number that $r$ sends to $1$ too, but it is not in the restricted domain, so it is not the value of THIS inverse.

Each point $(a, b)$ of $f$ becomes $(b, a)$: $(0, -2)$, $(1, -1)$, $(2, 1)$ become $(-2, 0)$, $(-1, 1)$, $(1, 2)$, drawn in the solution figure on equal scales. The graph of $f^{-1}$ crosses the $x$-axis at $(-2, 0)$, the image of the $y$-intercept $(0, -2)$ of $f$; it crosses the $y$-axis at $(0, \log_2 3)$, the image of the $x$-intercept of $f$, where $2^x = 3$. So $\log_2 3 = \frac{\ln 3}{\ln 2} \approx 1.5850$. Intercepts swap roles, exactly like domain and range.