over which of the following domains is f(x) = csc(0.1x + 1.2) defined at all points and invertible?

x = [0,10]
x = [10,20]
x = [20,30]
x = [30,40]

Can someone please explain how to solve this to me and show me all the steps rather than simply giving me the answer so I can understand it?

1. C (28.3)

2. E (sec x)
3. B (tan x and sec x)
4. E (sec x)
5. A (2.5 seconds)
6. D (a=1; b=-1; c=pi)
7. C (f(x)=2sin(-1.5x+0.5)-2)
8. B (theta(t)=6cos(2pi/3 t)
9. C (x=[20,30])
10. A (a decreasing function defined in Quadrants I and II)
11. B (51°)
12. D (sqrt 1-x^2)
13. E (arccos (cosx)=x)
14. D (10.6 mins)

the domain must be at most 1/2 period, or 2π/.1 = 20π. In addition, it must not contain an asymptote or a max/min, because there the curve doubles back, so it fails the horizontal-line test.

All of those intervals are short enough, so we need to consider that there is an asymptote at x = 10π-12 ≈ 19.4

there is a min/max where .1x+1.2=π/2 or 3π/2
x ≈ 4 and 34

Looks like [20,30] is the answer.

honors student is right but 13 is actually D. (arc cos (cosx)=x)

yeah pig benis is right

Well, let me break it down for you, but I must warn you, I like to throw in some humor along the way!

First, we need to determine over which domain the function f(x) = csc(0.1x + 1.2) is defined at all points. The cosecant function (csc) is defined as 1/sin(x), so it will be undefined whenever the sine function is zero.

To find where sin(x) is zero, we need to set 0.1x + 1.2 equal to multiples of π (since sin(x) repeats every π radians).

Are you ready for some math clowning around? Let's go!

1. Start with 0.1x + 1.2 = 0π:
-0.1x = 1.2 * 0π
-x = 0
-x = 0 * 0 (Multiplying anything by zero is still zero, even for mathematicians!)
-x = 0

2. Solve for x:
We found that x equals zero, which means sin(0) is zero.

Now, let's consider the given intervals:
a) x = [0,10]
b) x = [10,20]
c) x = [20,30]
d) x = [30,40]

Since sin(x) is zero at x = 0, we need to check if this value falls within any of the given intervals.

Looking at the intervals, we see that the first one, x = [0,10], includes x = 0. So, this is the correct domain for which the function is defined at all points.

But our circus fun doesn't stop here! We also need to check if the function is invertible. For a function to be invertible, it must pass the horizontal line test. In simpler terms, it means that no two points on the graph of the function can have the same y-coordinate.

Since f(x) = csc(0.1x + 1.2) is defined for all points in the interval x = [0,10], it means that the function is also invertible over this interval.

So, the answer is x = [0,10].

I hope this explanation brought some laughter along with the understanding!

To determine over which domain the function f(x) = csc(0.1x + 1.2) is defined at all points and invertible, we need to consider two conditions:

1. The domain should not include any values that make the denominator of csc(0.1x + 1.2) equal to zero. Since csc(x) is equivalent to 1/sin(x), it is undefined when sin(x) = 0.

The equation sin(x) = 0 has solutions at x = 0, π, 2π, 3π, and so on. These are known as the "zeroes" or "roots" of the sine function.

2. The function f(x) should also be injective or one-to-one, meaning that every input has a unique output.

Now, let's apply these conditions to the given options and solve step by step:

Option x = [0, 10]:
a) Check if there are any values within the interval that make the denominator zero:
0.1x + 1.2 = 0
0.1x = -1.2
x = -12

Since the interval does not include x = -12, the denominator will never be zero within the given range.

b) Check if the function is a one-to-one or injective function within the interval. To verify this, we observe that the function f(x) = csc(0.1x + 1.2) is periodic with a period of 2π/0.1 = 20π. Within the interval [0, 10] (which is smaller than 20π), the function does not repeat itself and is therefore injective. So, this interval satisfies the injective condition.

Option x = [10, 20]:
a) Check if there are any values within the interval that make the denominator zero:
0.1x + 1.2 = 0
0.1x = -1.2
x = -12

Again, x = -12 is not within the given interval, so the denominator will not be zero.

b) Check if the function is injective over this interval. Since the interval is a subset of the interval [0, 10], the function is still injective over this interval.

Option x = [20, 30]:
a) Check if there are any values within the interval that make the denominator zero:
0.1x + 1.2 = 0
0.1x = -1.2
x = -12

Similar to the previous options, x = -12 is not within the given interval, so the denominator will not be zero.

b) Check if the function is injective over this interval. Again, since the interval lies within the interval [0, 10], the function is injective over this interval as well.

Option x = [30, 40]:
a) Check if there are any values within the interval that make the denominator zero:
0.1x + 1.2 = 0
0.1x = -1.2
x = -12

Once again, x = -12 is not within the given interval, so the denominator will not be zero.

b) Check if the function is injective over this interval. Just like the previous options, since the interval is a subset of the interval [0, 10], the function is injective over this interval too.

Based on these analyses, we can see that for all the given options [0, 10], [10, 20], [20, 30], and [30, 40], the function f(x) = csc(0.1x + 1.2) is defined at all points and invertible.