Provide an example of a function that does not have an inverse function. How do you prove that it is not a function?

1 answer:

5 0

If f had an inverse, then its graph would be the reflection of the graph of f about the line y = x. The graph of f and its reflection about y = x are drawn below. Note that the reflected graph does not pass the vertical line test, so it is not the graph of a function.

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Jacques needs to convert the function f(x) = x2 − 6x + 14 to vertex form, f(x) = (x − h)2 + k, in order to find the minimum.

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k = 5

the equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k ) are the coordinates of the vertex and a is a multiplier

To obtain this form use the method of completing the square

Since the coefficient of the x² term is 1 then

add/ subtract (half the coefficient of the x-term )² to x² - 6x

f(x) = x² + 2(- 3)x + 9 - 9 + 14 = (x - 3)² + 5 → k = 5

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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