MAT 180: The SandBox

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91 thoughts on “MAT 180: The SandBox”

1. Anonymous says:

How do you solve problems like 10 on WeBWorK 14?

1. To solve an equation of the form
$$a \sin(bx -c) + d = 0$$
(assuming $$a \not = 0$$) isolate sine, use inverse sine, and solve for $$x$$.

Isolating sine we get,
$$\sin(bx -c) = -\frac{d}{a}$$

If we can use the unit circle to find an angle for which sine of that angle is equal to $$-\frac{d}{a},$$ then we don’t need to use inverse sine. If not use inverse sine to obtain
$$bx -c = \arcsin \left( – \frac{d}{a}\right)$$
Arcsine will only output one angle in the interval $$-\pi/2$$ to $$\pi/2$$, inclusive, for which sine of that angle is $$-\frac{d}{a}$$. If $$-\frac{d}{a} \geq 0$$ then arcsine outputs an angle in the first quadrant. The other angle for which sine is equal to $$-\frac{d}{a}$$ is obtained by taking $$\pi – \arcsin(-d/a)$$. Adding rotations to either of these angles will also be an angle for which sine of that angle is $$-\frac{d}{a}$$. If $$-\frac{d}{a} < 0$$ then arcsine outputs an angle in the fourth quadrant. The other angle for which sine is equal to $$-\frac{d}{a}$$ is obtained by taking $$-\pi + \arcsin(-d/a)$$. Adding rotations to either of these angles will also be an angle for which sine of that angle is $$-\frac{d}{a}$$. Now for an example with specific values of $$a,b,c,$$ and $$d$$.

Solve
$$2\sin(3x -1) + 1 = 0$$
(We’ll be able to use the unit circle).

Isolate sine:
$$\sin(3x-1) = – \frac{1}{2}$$
Sine is equal to $$-1/2$$ at $$-\pi/6.$$ The other angle for which sine is equal to $$-1/2$$ is in the third quadrant and is equal to $$-\pi + \pi/6 = -5 \pi/ 6$$. To find all solutions we need to allow for rotations of these two solutions. So,
$$3x -1 = \frac{-\pi}{6} + 2 \pi n$$
or
$$3x -1 = \frac{-5 \pi}{6} + 2 \pi n$$

Finally, solving for $$x$$ we obtain:

$$x = \frac{-\pi}{18}+\frac{1}{3} + \frac{2 \pi}{3} n$$
or
$$x = \frac{-5 \pi}{18} +\frac{1}{3} + \frac{2 \pi}{3} n$$

If you wanted the smallest positive value, choose the smallest $$n$$ such that $$x$$ is positive

Solve
$$10\sin(3x -1) – 1 = 0$$
(We’ll need to use inverse sine).

Isolate sine:
$$\sin(3x-1) = \frac{1}{10}$$

Using inverse sine:
$$3x -1 = \arcsin \left( \frac{1}{10} \right)$$

The other angle for which sine is equal to $$1/10$$ is in the second quadrant and is equal to $$\pi – \arcsin(1/10)$$. To find all solutions we need to allow for rotations of these two solutions. So,
$$3x -1 = \arcsin(1/10) + 2 \pi n$$
or
$$3x -1 = \pi-\arcsin(1/10) + 2 \pi n$$

Finally, solving for $$x$$ we obtain:

$$x = \frac{\arcsin(1/10)}{3} + \frac{1}{3} + \frac{2 \pi}{3} n$$
or
$$x = \frac{\pi – \arcsin(1/10)}{3} + \frac{1}{3} + \frac{2 \pi}{3} n$$

If you wanted the smallest positive value, choose the smallest $$n$$ such that $$x$$ is positive. In this case, $$n = 0$$ and $$x = \frac{\arcsin(1/10)}{3} + \frac{1}{3}$$.

2. Anonymous says:

I have a problem with Homework 13 Problem 2… we have overlapping sets of x like: 4<x<12 and 8<=x<=10. Which function should we use? Or we should graph results for both?

1. That’s a typo. $$\log_2(x-4)$$ should be defined on the interval $$4$$ to $$8$$, not inclusive.

I’ve edited the file to reflect these changes.

3. Anonymous says:

Professor Nevo, I’m not sure how to input my answer for Problem 7 on WeBWorK 13. My solution sets are x>=4 and x>=-4, but negatives can’t work because of the function being under a radical. 0 seems to work as well as all positives. How would I input this using interval notation?

1. You may have solved the corresponding inequality incorrectly. Make sure you graph the function underneath the square root, then identify the intervals along the $$x$$-axis for which the function is above or on the $$x$$-axis. This will be the solution set for the inequality that corresponds to the domain of the function.

2. Anonymous says:

Okay I graphed it. It’s vertex is at (0,-16) with x intercepts at (-4,0) and (4,0). The parabola opens upward given a>0. Now the solution sets would have to be x=4. Right?

3. According to what you described the solution set should be all $$x$$ values to the left of $$-4$$ and to the right of $$4$$, including $$\pm 4$$. In interval notation, this would be
$$(-\infty, -4] \cup [4, \infty)$$

But I thought that problem 7 has a rational function under the radical, no? If it does, your graph should be that of a rational function, not a parabola.

4. Anonymous says:

What’s the beginning steps to solving problem 1 on Webwork 12?

1. Let the left-hand side of the inequality be the function $$r(x)$$. The first step should be to graph $$r$$.

5. Anonymous says:

Professor i got a parabola graph on the problem no. 2 so my solution would be in between of the two endpoint but when i put it in it says incorrect

1. Are you supposed to include both the endpoints, one of the endpoints, or none of the endpoints. If you need to include an endpoint in your solution then you use a bracket instead of a parenthesis.

6. Anonymous says:

For WeBWorK 12, Problem 2, x can never be less than zero. How would I input that into WeBWorK?

1. If $$x$$ cannot be less than zero, then $$x$$ must be greater than or equal to zero. That is, $$x \geq 0$$. In interval notation that is $$[0, \infty)$$, where WeBWorK uses Inf for $$\infty$$.

7. Anonymous says:

How is (-infinity,3]U[9,infinity) not the answer for number 4?

1. Consider the inequality
$$\frac{3-x}{x-8} \geq 0$$

Graph $$f(x) = \frac{3-x}{x-8}$$.

The graph of $$f$$ is above the $$x$$-axis for $$x$$ values between 3 and 8, including 3 but excluding 8. That is,
$$\frac{3-x}{x-8} \geq 0$$
on the interval $$[3, 8)$$.

8. Anonymous says:

How do we answer question #7 for the non-vertical asymptote? We divide the denominator by the numerator, correct? How do we make -7x^2 into 5x^3?

1. You divide the numerator by the denominator. The quotient will then be the slant (or non-vertical) asymtote.

9. Anonymous says:

When did we learn about non-vertical asymptotes or how to answer the last two questions? I don’t recall learning that at all on Thursday.

1. The last two questions can be solved using similar techniques to solving for constants in formulas for exponential, trigonometric, or logarithmic functions. Plug in some values for x and see what the constants values have to be in order for the equation to make sense.