A function does not have any x-intercepts. What might be true about its domain and range?

Mathematics

1 answer:

timama [110]2 years ago

6 0

Answer:

Here's my two cents worth about this one:

This is a discontinuous function - and it passes the vetical line test because any vertical line would only cut the "graph" in one place. The graph would have an infinite number of disconnected "peaks" and "valleys" because the x values along the number line are continuous. In other words, all the rationals would have y values of 1 and all the irrationals would have values of 0. The graph is definitely non-continuous because there aren't any "intermediate" y values - i.e., "y" is either 0 or 1 - and nothing in between !! Basically, the graph would be just "dots" at either y= 0 or at y =1. In a strange way though, this is an "even" function, because the negative of any positive rational would have a corresponding y value of 1 and the negative of any positive irrational would have a corresponding y value of 0. Thus, f(x) = f(-x). The y intercept would be the "dot" at (0, 1) - because 0 is rational. The "x" intercepts would just be the "dots" where the irrational numbers are located on the number line because "f(x)" would be 0 at those points!!! The domain is all real numbers, but there are onl two values for the "range" .....0 and 1 !!!

That's my "take" on this one !!!

Step-by-step explanation:

You might be interested in

What is 8,100 in scientific notation?

Rzqust [24]

8.1 I think but there's an app called, tiger something and it tells,you the answers

6 0

3 years ago

What are the solutions of the quadratic equation (x + 3)2 = 49

Korvikt [17]

the right answer is

x = 4 and x = –10

I just take the test and I got it right I hope yall get it right to

8 0

3 years ago

Read 2 more answers

A researcher claims that the variation in the salaries of elementary school teachers is greater than the variation in the salari

guajiro [1.7K]

Answer:

The null hypothesis is $H_o : \sigma^2_1 = \sigma^2 _2$

The alternative hypothesis is $H_a : \sigma_1 ^2 > \sigma^2_2$

$F_{critical} = 1.8608$

$F = 2.9085$

The decision rule is

Reject the null hypothesis

There is sufficient evidence to support the researchers claim

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 30$

The sample variance for elementary school is $s^2_1 = 8324$

The second sample size is $n_2 = 30$

The sample variance for the secondary school is $s^2_2 = 2862$

The significance level is $\alpha = 0.05$

The null hypothesis is $H_o : \sigma^2_1 = \sigma^2 _2$

The alternative hypothesis is $H_a : \sigma_1 ^2 > \sigma^2_2$

Generally from the F statistics table the critical value of $\alpha = 0.05$ at first and second degree of freedom $df_1 = n_1 - 1 = 30 - 1 = 29$ and $df_2 = n_2 - 1 = 30 - 1 = 29$ is