ap precalculus - mixed six - unit 3 review\nno calculator\nthree functions f, g, and h are introduced in #1…

Explanation:

Step1: Recall vertical - asymptote formula for tangent function

The vertical asymptotes of (y = \tan(kx)) occur at (kx=\frac{\pi}{2}+n\pi), (n\in\mathbb{Z}). For (f(x)=\tan(\frac{1}{2}x)+4), we set (\frac{1}{2}x=\frac{\pi}{2}+n\pi). Solving for (x) gives (x=\pi + 2n\pi). In the interval ((0,2\pi)), when (n = 0), (x=\pi).

Step2: Solve for input when output is 4

Set (f(x)=4), so (\tan(\frac{1}{2}x)+4 = 4), then (\tan(\frac{1}{2}x)=0). The solutions of (\tan t = 0) are (t=n\pi), (n\in\mathbb{Z}). So (\frac{1}{2}x=n\pi), and (x = 2n\pi). In the interval ((0,2\pi)), when (n = 1), (x = 2\pi) (but this is not in the open - interval ((0,2\pi)), when (n=0), there is no solution in the open interval ((0,2\pi))).

Answer:

a) (x=\pi) b) No solution in ((0,2\pi))

Explanation:

Step1: Calculate amplitude of (g(x))

The amplitude (A) of a sinusoidal function (y = A\sin(bx + c)+d) is given by (A=\frac{\text{Max}-\text{Min}}{2}). Given (\text{Max}(g)=8) and (\text{Min}(g)=- 2), then (A=\frac{8 - (-2)}{2}=\frac{10}{2}=5).

Step2: Find the period of (g(x))

The period of a sinusoidal function can be found by looking at the (x) - values where the function repeats. From the table, the function (g(x)) has a pattern that repeats every (6) units of (x) (e.g., from (x=-4) to (x = 2), (g(-4)=8) and (g(2)= - 2), and from (x = 2) to (x=8), the pattern repeats). So the period (T = 6).

Answer:

a) (5) b) (6)

Explanation:

Step1: Recall amplitude formula for (y = A\cos(Bx + C))

For the function (h(x)=3\cos(2(x+\frac{\pi}{8}))), the amplitude (A) is the absolute value of the coefficient of the cosine function. So (A = 3), not (2), so the statement "The amplitude of the graph of (h) is (2)" is False.

Step2: Recall period formula for (y = A\cos(Bx + C))

The period (T) of (y=\cos(Bx)) is (T=\frac{2\pi}{B}). For (h(x)=3\cos(2(x+\frac{\pi}{8}))), (B = 2), so (T=\frac{2\pi}{2}=\pi), the statement "The period of the graph of (h) is (\pi)" is True.

Step3: Recall phase - shift formula for (y = A\cos(Bx + C))

The phase - shift of (y = A\cos(Bx + C)) is (-\frac{C}{B}). For (h(x)=3\cos(2(x+\frac{\pi}{8}))=3\cos(2x+\frac{\pi}{4})), the phase - shift is (-\frac{\pi}{8}) (a left - shift of (\frac{\pi}{8}) units), so the statement "The horizontal shift (phase shift) of the graph of (h) is right (\frac{\pi}{8}) units" is False.

Step4: Recall range formula for (y = A\cos(Bx + C))

The range of (y = A\cos(Bx + C)) is ([-A,A]). For (h(x)=3\cos(2(x+\frac{\pi}{8}))), (A = 3), so the range is ([-3,3]), the statement "The range of the graph of (h) is ([-3,3])" is True.

Answer:

The amplitude of the graph of (h) is (2): False The period of the graph of (h) is (\pi): True The horizontal shift (phase shift) of the graph of (h) is right (\frac{\pi}{8}) units: False The range of the graph of (h) is ([-3,3]): True

Explanation:

The function (y = \tan x) is transformed to (f(x)=\tan(\frac{1}{2}x)+4). The transformation (y=\tan x\to y = \tan(\frac{1}{2}x)) is a horizontal stretch by a factor of (2) (since the period of (y=\tan x) is (\pi) and the period of (y=\tan(\frac{1}{2}x)) is (2\pi)), and the (+4) is a vertical shift of (4) units up. To sketch the graph of (f(x)) on ([-2\pi,2\pi]), we first mark the vertical asymptotes of (y=\tan(\frac{1}{2}x)) which are (x=\pi+2n\pi). In the interval ([-2\pi,2\pi]), the vertical asymptotes are (x =-\pi) and (x=\pi). Then we can plot some key points of (y = \tan(\frac{1}{2}x)) (e.g., when (x = 0), (y = 0), then shift these points up by (4) units.

Answer:

Explanation of transformations: Horizontal stretch by a factor of 2 and vertical shift 4 units up. Sketch: Mark asymptotes (x =-\pi) and (x=\pi) in ([-2\pi,2\pi]), plot key - points of (y=\tan(\frac{1}{2}x)) and shift up 4 units.

Explanation:

Step1: Find (a)

The amplitude (a) of (g(x)=a\sin(b(x + c))+d) is (a = 5) (from part 2a).

Step2: Find (b)

The period (T=\frac{2\pi}{b}), and since (T = 6), then (b=\frac{2\pi}{6}=\frac{\pi}{3}).

Step3: Find (c)

We can use a point - by - point analysis. Let's assume the function is of the form (y=a\sin(bx+bc)+d). We know the mid - line (y = d=\frac{8+( - 2)}{2}=3). Let's use the fact that the function has a maximum at some point. If we assume the standard form of a sine function (y = a\sin(bx + c)+d) and we know (a = 5), (b=\frac{\pi}{3}), (d = 3). By looking at the table and the general form of the sine function, we can find (c). If we consider the phase - shift, and assume the function starts its cycle at a known point. Let (x = 2) be a point where the function is at a minimum. For (y = 5\sin(\frac{\pi}{3}(x + c))+3), when (x = 2), (y=-2). Then (-2=5\sin(\frac{\pi}{3}(2 + c))+3), (\sin(\frac{\pi}{3}(2 + c))=-1), (\frac{\pi}{3}(2 + c)=\frac{3\pi}{2}+2n\pi), (2 + c=\frac{9}{2}+6n), (c=\frac{5}{2}+6n). Let (n = 0), (c=\frac{5}{2}).

Step4: Find (d)

The mid - line of the sinusoidal function (g(x)) is (d=\frac{\text{Max}+\text{Min}}{2}=\frac{8+( - 2)}{2}=3).

Answer:

(a = 5) (b=\frac{\pi}{3}) (c=\frac{5}{2}) (d = 3)

Explanation:

The function (h(x)=3\cos(2(x+\frac{\pi}{8}))) and (j(x)=3\sec(2(x+\frac{\pi}{8}))=\frac{3}{\cos(2(x+\frac{\pi}{8}))}). The vertical asymptotes of (j(x)) occur where (\cos(2(x+\frac{\pi}{8})) = 0). So (2(x+\frac{\pi}{8})=\frac{\pi}{2}+n\pi), (2x+\frac{\pi}{4}=\frac{\pi}{2}+n\pi), (2x=\frac{\pi}{4}+n\pi), (x=\frac{\pi}{8}+\frac{n\pi}{2}). In the interval ([0,\pi]), when (n = 0), (x=\frac{\pi}{8}), when (n = 1), (x=\frac{5\pi}{8}). Susan is not correct.

Answer:

Susan is not correct. The vertical asymptotes of (j(x)) in ([0,\pi]) are (x=\frac{\pi}{8}) and (x=\frac{5\pi}{8}).