Practice Final
Math74 Completion of Answer Key

8.

$$\left(-\frac{1}{27}\right)^{-4/3} = 81$$

26.

The solution set is the set of all real numbers. To graph it, you would shade in the entire real line.

(Reason: The solution sets to each of the given inequalities are $x\leq -2$ and $x\geq -4$, respectively. Together, these two intervals include every real number.)

42.

(This is a circle of radius 2, with center (1,0).)

44.

(This is a parabola with vertex (3,-4) and axis of symmetry x=3. It opens downward, so there are no x-intercepts.)

51.

$$x\geq -5$$ Domain:

$$y\geq 0$$ Range:

$$f^{-1}(x) = x^2-5$$ Inverse:

$$\mbox{Domain of $f^{-1}$: } x\geq 0$$

$$\mbox{Range of $f^{-1}$: } y\geq -5$$

(Note that the range of $f^{-1}$ is the domain of $f$, and the domain of $f^{-1}$ is the range of $f$.)

Graphs:

52.

Domain: All real numbers.

$$y\geq 0$$ Range:

Inverse: None, since the function is not one-to-one.

53.

Horizontal asymptote: $y=-\frac{10}{7}$.

54.

Vertical asymptotes: $x=\pm\sqrt{\frac{2}{3}}$.

55.

In the following graph, the point (1,5) is included (closed circle), but the point (1,-2) is not (open circle).

56.

59.

(-4,-5) is also on (the graph of) P(x).

64.

65.

$\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$

66.

$s=\frac{7}{6}\pi + 2\pi n,$ where n = any integer.

67.

If $b\neq 0$ and $-1<b<1$, then there are two solutions.
If $b=0$, there are three solutions: $t=0,\pi,2\pi$.
If $b=1$, there is one solution: $t=\pi/2$.
If $b=-1$, there is one solution: $t=3\pi/2$.

68.

This problem is ambiguous, since we're not given the quadrant for $t$. So there will be two possible answers.

If $\cos (-t) =0.2$, then $\cos t = 0.2$, and so $\sin t = \pm\sqrt{1-\cos^2 t} = \pm\sqrt{0.96}.$ Since $\sin(t+31\pi)=-\sin t$,

$$\sin(t+31\pi) = \pm\sqrt{0.96}.$$

(If $t$ is in quadrant I or II, then $\sin(t+31\pi)=-\sqrt{0.96};$ if $t$ is in quadrant III or IV, $\sin(t+31\pi)=\sqrt{0.96}.$

69.

The amplitude is 2, and the period is 8.

70.

71.

72.

As far as I can tell, this is not possible...

73.

Solution: 1/2.

89.

I guess they're asking for something like this:

Sum identity:

$$\cos(120 ^{\circ}) = \cos(60^{\circ}+60^{\circ})$$

$$= \cos(60^{\circ})\cos(60^{\circ}) - \sin(60^{\circ})\sin(60^{\circ})$$

$$= \left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2$$

$$= \frac{1}{4}-\frac{3}{4} = -\frac{1}{2}$$

Difference identity:

$$\cos(120 ^{\circ}) = \cos(180^{\circ}-60^{\circ})$$

$$= \cos(180^{\circ})\cos(60^{\circ}) + \sin(180^{\circ})\sin(60^{\circ})$$

$$= (-1)\frac{1}{2} + 0(\frac{\sqrt{3}}{2})$$

$$= -\frac{1}{2}$$