Chapter 1 Solutions

Authored worked explanation for the original source problem.

Source review note: The source short answer is 2. The printed expression includes a final −1, so the reviewed result is 1.

1. At $5\pi /4$, $\sin (5\pi /4)=-\sqrt{2}/2$ and $\tan (5\pi /4)=1$.

   Why: Both coordinates are negative in quadrant III, so tangent is positive.

2. Thus $\csc (5\pi /4)=-\sqrt{2}$, and $(-\sqrt{2}·1{)}^2-1=2-1=1$.

   Why: Take the reciprocal of sine before squaring the product.

Reviewed answer: 1

Takeaway: Squaring removes a sign; it does not remove the final subtraction.

Authored worked explanation for the original source problem.

1. $\cot (\pi /6)=\sqrt{3}$ and $\csc (\pi /6)=2$.

   Why: Use the 30–60–90 triangle.

2. $(\sec a+\tan a)(\sec a-\tan a)={\sec }^{2}a-{\tan }^{2}a=1$ with $a=\pi /8$.

   Why: The denominator is a difference of squares and the secant identity makes it 1.

3. The quotient is $2\sqrt{3}/1=2\sqrt{3}$.

   Why: There is no need to approximate the π/8 values.

Reviewed answer: $2\sqrt{3}$

Takeaway: Simplify an identity before reaching for a calculator.

Authored worked explanation for the original source problem.

1. At $3\pi /4$, $\csc =\sqrt{2}$, $\tan =-1$, $\sec =-\sqrt{2}$, and $\cot =-1$.

   Why: Sine is positive and cosine is negative in quadrant II.

2. The two squared products are $(-\sqrt{2}{)}^2=2$ and $(\sqrt{2}{)}^2=2$.

   Why: Square each entire product, then add.

3. $2+2=4$.

   Why: Both terms are nonnegative after squaring.

Reviewed answer: $4$

Takeaway: Track the signs before you square them away.

Authored worked explanation for the original source problem.

1. The graph has maximum 4 and minimum −2, so amplitude is $(4-(-2\left)\right)/2=3$ and midline is $(4+(-2\left)\right)/2=1$.

   Why: Amplitude is half the vertical range; the midline is its average.

2. Successive peaks are $4\pi$ apart, so the horizontal coefficient is $2\pi /\left(4\pi \right)=1/2$.

   Why: The period of sin(bx) is 2π/|b|.

3. The curve rises through its midline at $x=\pi$, giving $y=3\sin \left(\right(x-\pi )/2)+1$.

   Why: An increasing midline crossing is the convenient zero phase for a positive sine wave.

Reviewed answer: $y=3\sin \left(\right(x-\pi )/2)+1$

Takeaway: Read height, period, and phase as three separate features.

Authored worked explanation for the original source problem.

1. From problem 4, the sine angle is $x/2-\pi /2$.

   Why: Keep the same amplitude, midline, and period.

2. $\sin (u-\pi /2)=-\cos u$, so $3\sin (x/2-\pi /2)+1=-3\cos (x/2)+1$.

   Why: A phase identity converts the same curve from sine to cosine.

3. Equivalently $y=3\cos \left(\right(x-2\pi )/2)+1$; adding another full period gives the other source form.

   Why: A phase shift by a whole period does not change a periodic graph.

Reviewed answer: $y=-3\cos (x/2)+1$

Takeaway: Different-looking formulas can describe the same sinusoid.

Authored worked explanation for the original source problem.

Source review note: The source gives (1/2)e^(−x)+1, which reflects rather than shifts right. The reviewed formula uses the usual combined-transform convention; literally applying the listed up-shift before compression would instead give (1/2)e^(x−1)+1/2, so the order wording is also ambiguous.

1. Replace $x$ by $x-1$ to get $e^{x-1}$.

   Why: Inside subtraction shifts the graph one unit right.

2. Multiply the exponential height by 1/2 and put the final midline at 1: $g\left(x\right)=(1/2)e^{x-1}+1$.

   Why: This uses the standard transform convention: compress relative to the original x-axis, then translate upward.

Reviewed answer: $g\left(x\right)=(1/2)e^{x-1}+1$

Takeaway: Inside changes move input locations; outside changes alter heights.

Authored worked explanation for the original source problem.

Source review note: The source lists domain (−1,1) and range all real numbers. Principal arcsin includes ±1 and has range [−π/2,π/2].

1. Sine restricted to $[-\pi /2,\pi /2]$ is one-to-one and reaches every value in $[-1,1]$.

   Why: A function must be one-to-one before it has an inverse function.

2. Swap the restricted sine domain and range: arcsin accepts inputs from −1 to 1 and returns angles from $-\pi /2$ to $\pi /2$, including the endpoints.

   Why: The inverse exchanges inputs with outputs.

Reviewed answer: Domain [−1,1]; range [−π/2,π/2].

Takeaway: An inverse trig function returns one principal angle, not every possible angle.

Authored worked explanation for the original source problem.

Source review note: The source reports 3.36; the value is about 3.36509, which rounds to 3.37 to two decimal places. The source is retained.

1. ${\log }_{4}20=\ln \left(20\right)/\ln \left(4\right)$ and ${\log }_{10}16=\ln \left(16\right)/\ln \left(10\right)$.

   Why: Change of base replaces each logarithm with a natural logarithm quotient.

2. Move constant factors into powers, then combine the sum: $y=\ln \left({20}^{1/\ln 4}\right)+\ln \left({16}^{1/\ln 10}\right)=\ln ({20}^{1/\ln 4}·{16}^{1/\ln 10})$.

   Why: The power and product log rules apply because both arguments are positive.

3. Evaluate the two quotients to get approximately 3.3651.

   Why: Keep extra digits until the end; the source's 3.36 is a truncated approximation.

Reviewed answer: $y=\ln ({20}^{1/\ln 4}·{16}^{1/\ln 10})\approx 3.3651$

Takeaway: A sum of logarithms becomes the log of a product.

Authored worked explanation for the original source problem.

1. $\ln \left(\right(3e/\pi {)}^2)=2\ln (3e/\pi )$.

   Why: The argument is positive, so the power rule applies.

2. Expand to $2\ln 3+2\ln e-2\ln \pi$ and use $\ln e=1$.

   Why: Products add logs and quotients subtract logs.

3. $2\ln 3=\ln 9$, giving $2+\ln 9-2\ln \pi$.

   Why: This matches the requested form with single factors inside each logarithm.

Reviewed answer: $2+\ln 9-2\ln \pi$

Takeaway: Expand the argument's multiplication and powers, not a sum inside a logarithm.

Authored worked explanation for the original source problem.

Source review note: The source key says “Nope”. These authored formulas were inferred from the source raster with its labeled axes; exact generating equations are not embedded in the archive.

1. The green curve passes through $(0,1)$ and decreases toward the x-axis; blue passes through $(0,-1)$ and falls to the right; orange passes through $(0,-1)$ and rises toward zero to the right.

   Why: Reflecting the black exponential across the y-axis gives e^(−x); reflecting across the x-axis gives −e^x; both reflections give −e^(−x).

2. Yellow passes through $(1,0)$, is increasing, and has a vertical asymptote at x=0, so it matches $\ln x$. Purple is its reflection $-\ln x$.

   Why: Swapping exponential inputs and outputs yields its inverse logarithm; the intercept and asymptote distinguish it from an exponential transform.

3. Cyan and pink are the same log pair reflected into x<0, giving $\ln (-x)$ and $-\ln (-x)$. They pass through $(-1,0)$.

   Why: Replacing x with −x reflects a graph across the y-axis and changes its domain to negative inputs.

4. Check the inferred equations against several grid landmarks, such as the green height at x=1 and the yellow height at x=2.

   Why: The source is a raster plot, so these equations are inferred from its labeled scale rather than recovered from original generating data.

Reviewed answer: Green $e^{-x}$; blue $-e^x$; orange $-e^{-x}$; pink $-\ln (-x)$; cyan $\ln (-x)$; purple $-\ln x$; yellow $\ln x$.

Takeaway: Reflections explain the exponentials; inverse functions explain the logarithms.

Authored solution to an added practice problem.

1. $11\pi /6$ lies in quadrant IV, with $\cos =\sqrt{3}/2$ and $\sin =-1/2$.

   Why: The reference angle is π/6.

2. $\sin (11\pi /6)·\sec (11\pi /6)=(-1/2)(2/\sqrt{3})=-\sqrt{3}/3$.

   Why: Secant is the reciprocal of cosine.

Reviewed answer: $-\sqrt{3}/3$

Takeaway: Use the reference angle and quadrant together.

Authored solution to an added practice problem.

1. $\tan \theta ·\cot \theta =(\sin \theta /\cos \theta )(\cos \theta /\sin \theta )=1$.

   Why: The reciprocal factors cancel only when their original denominators are nonzero.

2. Therefore exclude multiples of π/2, where sine or cosine vanishes.

   Why: Simplification must not add points to the original domain.

Reviewed answer: $1$, where $\sin \theta \ne 0$ and $\cos \theta \ne 0$.

Takeaway: A simplified constant can still have domain restrictions.

Authored solution to an added practice problem.

1. $2\sin x=1$ means $\sin x=1/2$.

   Why: Isolate the trig function first.

2. On $0\le x<2\pi$, sine is 1/2 in quadrants I and II: $\pi /6$ and $5\pi /6$.

   Why: Both angles have the same positive vertical coordinate on the unit circle.

Reviewed answer: $x=\pi /6$ or $x=5\pi /6$.

Takeaway: An equation may have more solutions than the principal inverse value.

Authored solution to an added practice problem.

1. Since θ is in quadrant II, $\cos \theta =-\sqrt{1-(3/5{)}^2}=-4/5$.

   Why: The quadrant determines the missing square-root sign.

2. $\sin \left(2\theta \right)=2\sin \theta \cos \theta =2(3/5)(-4/5)=-24/25$.

   Why: Use the double-angle identity without finding θ numerically.

Reviewed answer: $\sin \left(2\theta \right)=-24/25$.

Takeaway: The sign of the missing trig value matters.

Authored solution to an added practice problem.

1. Write the angle as $3(x-\pi /6)$.

   Why: Factor the input coefficient before reading the horizontal shift.

2. The outside coefficient −2 gives amplitude 2 and reflection; the period is $2\pi /3$, and −4 is the midline.

   Why: Amplitude uses absolute value, while a negative sign reverses peaks and troughs.

Reviewed answer: Amplitude 2; period $2\pi /3$; midline $y=-4$; shift $\pi /6$ right.

Takeaway: Read a horizontal shift only after factoring the angle.

Authored solution to an added practice problem.

1. The amplitude is $(3-(-5\left)\right)/2=4$ and the midline is $(3+(-5\left)\right)/2=-1$.

   Why: Use the highest and lowest values.

2. Peaks six units apart give $b=2\pi /6=\pi /3$; a peak at zero makes cosine convenient.

   Why: Positive cosine begins at a maximum.

3. Combine the features: $y=4\cos (\pi x/3)-1$.

   Why: The equation then has peaks at 0 and 6 as required.

Reviewed answer: $y=4\cos (\pi x/3)-1$.

Takeaway: Choose a base function that fits a visible landmark.

Authored solution to an added practice problem.

1. Moving $\sqrt{x}$ left three replaces x with $x+3$.

   Why: Inside addition shifts inputs left.

2. Apply reflection and stretching to get $-2\sqrt{x+3}$, then add 4.

   Why: The specified order leaves the final vertical shift outside the multiplication.

3. Require $x+3\ge 0$. The square root is nonnegative, so the output is at most 4.

   Why: The input radical and output sign determine the domain and range.

Reviewed answer: $g\left(x\right)=-2\sqrt{x+3}+4$; domain $[-3,\infty )$; range $(-\infty ,4]$.

Takeaway: Specify transformation order when outside operations do not commute.

Authored solution to an added practice problem.

1. For $g\left(x\right)=3f(x+1)-2$, x=−1 selects $f\left(0\right)=4$ and x=0 selects $f\left(1\right)=1$.

   Why: The transformed input tells you which table entry to use.

2. $g(-1)=3·4-2=10$ and $g\left(0\right)=3·1-2=1$.

   Why: Only after selecting the input do we apply the outside operations.

Reviewed answer: $g(-1)=10$ and $g\left(0\right)=1$.

Takeaway: An input shift changes the table lookup, not the table's heights.

Authored solution to an added practice problem.

1. Right two gives $e^{x-2}$; stretching by 3 and moving down one gives $3e^{x-2}-1$.

   Why: Each operation appears in its own place in the formula.

2. As $x\to -\infty$, the exponential tends to 0, so the horizontal asymptote is −1. At x=2 the exponential is 1, so $g\left(2\right)=2$.

   Why: A shift moves the exponential's baseline but not its positive nature.

Reviewed answer: $g\left(x\right)=3e^{x-2}-1$; asymptote $y=-1$; $g\left(2\right)=2$.

Takeaway: An asymptote and one landmark quickly check an exponential transform.

Authored solution to an added practice problem.

1. Separate the product and quotient: $\ln \left(x^2\right(x+1)/5)=\ln \left(x^2\right)+\ln (x+1)-\ln 5$.

   Why: All factors are positive on the stated domain.

2. Use $\ln \left(x^2\right)=2\ln x$ since x>0.

   Why: The positive-domain condition avoids an omitted absolute value.

Reviewed answer: $2\ln x+\ln (x+1)-\ln 5$, for $x>0$.

Takeaway: Log rules come with domain assumptions.

Authored solution to an added practice problem.

1. Move the coefficient: $2\ln x=\ln \left(x^2\right)$.

   Why: A log coefficient becomes a power.

2. Add the ln5 and subtract ln(x−1) to get $\ln (5x^2/(x-1\left)\right)$.

   Why: Adding multiplies arguments; subtracting divides them.

Reviewed answer: $\ln (5x^2/(x-1\left)\right)$, for $x>1$.

Takeaway: Preserve the original positive arguments when combining logs.

Authored solution to an added practice problem.

1. Combine to $\ln \left(\right(x-1\left)\right(x+1\left)\right)=\ln 8$, with $x>1$.

   Why: The original logarithms both need positive arguments.

2. $x^2-1=8$ gives $x=\pm 3$, but only 3 lies in the domain.

   Why: Exponentiating a log equation is valid on its domain; algebra alone can introduce an invalid candidate.

Reviewed answer: $x=3$.

Takeaway: Check proposed roots against the original logarithms.

Authored solution to an added practice problem.

1. ${\log }_{4}8=\ln 8/\ln 4$.

   Why: Change to a common base.

2. $\ln 8=3\ln 2$ and $\ln 4=2\ln 2$, so the ratio is 3/2.

   Why: Powers of the same positive base make the exact value visible.

Reviewed answer: $3/2$.

Takeaway: An exact value can be easier than a decimal.

Authored solution to an added practice problem.

1. Set $y=x^2+1$ with $x\ge 0$, swap x and y, and solve $x=y^2+1$.

   Why: An inverse reverses input and output roles.

2. The original nonnegative restriction chooses $y=\sqrt{x-1}$ rather than the negative branch.

   Why: The inverse range must equal the original domain.

3. The original range is $[1,\infty )$, so that becomes the inverse's domain.

   Why: Square roots require a nonnegative radicand.

Reviewed answer: $f^{-1}\left(x\right)=\sqrt{x-1}$, domain $[1,\infty )$.

Takeaway: The original domain restriction chooses the inverse branch.

Authored solution to an added practice problem.

1. $\sin (5\pi /6)=1/2$, and principal arcsin returns the angle in $[-\pi /2,\pi /2]$.

   Why: 5π/6 lies outside that output interval.

2. $5\pi /6$ lies in arccos's principal interval $[0,\pi ]$, so arccos returns the same angle.

   Why: The two inverse trig functions use different principal intervals.

Reviewed answer: $\arcsin \left(\sin \right(5\pi /6\left)\right)=\pi /6$, while $\arccos \left(\cos \right(5\pi /6\left)\right)=5\pi /6$.

Takeaway: Inverse trig only cancels trig within its chosen principal interval.

Authored solution to an added practice problem.

1. $\ln \left(e^x\right)=x$ because $e^x$ is always positive.

   Why: The input to ln is valid for every real x.

2. $e^{\ln x}=x$ requires x>0, the domain of ln.

   Why: The composition cannot be used where its inner function is undefined.

Reviewed answer: $f\left(g\right(x\left)\right)=x$ for every real x; $g\left(f\right(x\left)\right)=x$ only for $x>0$.

Takeaway: A composition inherits the inner function's domain.

Authored solution to an added practice problem.

1. Tangent is undefined when its angle is $\pi /2+k\pi$.

   Why: Its cosine denominator vanishes there.

2. For $\tan \left(2x\right)$, solve $2x=\pi /2+k\pi$ to get $x=\pi /4+k\pi /2$.

   Why: The input factor halves both the period and the spacing between vertical asymptotes.

3. Subtracting 1 moves the graph vertically but leaves those inputs unchanged.

   Why: A vertical translation cannot repair an undefined input.

Reviewed answer: Domain excludes $x=\pi /4+k\pi /2$ for integers k; period $\pi /2$.

Takeaway: Vertical movement does not shift vertical asymptotes horizontally.

Authored solution to an added practice problem.

1. At x=2, $g\left(2\right)=af\left(0\right)+b=a+b=3$.

   Why: The horizontal shift sends x=2 back to the table's input 0.

2. At x=3, $g\left(3\right)=af\left(1\right)+b=3a+b=-1$.

   Why: Use the second known height.

3. Subtract the equations: $2a=-4$, so a=−2 and b=5.

   Why: Two distinct source heights determine the vertical scale and translation.

Reviewed answer: $a=-2$, $b=5$.

Takeaway: A transformed graph's landmarks can solve for its unknown parameters.

Authored solution to an added practice problem.

1. The valid identity is $1+{\tan }^2\theta ={\sec }^2\theta$.

   Why: The square on secant comes from dividing sin²+cos²=1 by cos².

2. At π/4, $1+{\tan }^2\theta =2$, but $\sec \theta =\sqrt{2}$.

   Why: One valid counterexample disproves the proposed identity.

Reviewed answer: The claim is false; at $\theta =\pi /4$ the left side is 2 and ${\sec }^2\theta =2$, while $\sec \theta =\sqrt{2}$.

Takeaway: A missing square changes an identity into a false statement.

Authored solution to an added practice problem.

1. Set $y=2e^{x-3}+5$ and isolate $e^{x-3}=(y-5)/2$.

   Why: Undo the outside addition and multiplication first.

2. Take ln: $x=\ln \left(\right(y-5)/2)+3$; now interchange input and output names.

   Why: Ln undoes the positive exponential.

3. Since the original range is $(5,\infty )$, require the inverse input x>5.

   Why: The inverse domain is the original range.

Reviewed answer: $g^{-1}\left(x\right)=\ln \left(\right(x-5)/2)+3$ for $x>5$.

Takeaway: Undo a transform in reverse order.