# Let f(x)=4−frac{2}{x}+frac{6}{x^{2}}. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima). f is increasing on the intervals f is decreasing on the intervals The relative maxima of f occur at x = The relative minima of f occur at x = Notes: In the first two, your answer should either be a single interval, such as (0,1), a comma separated list of intervals, such as (-inf, 2), (3,4), or the word "none". In the last two, your answer should be a comma separated list of x values or the word "none".

Question
Confidence intervals
Let $$f(x)=4−\frac{2}{x}+\frac{6}{x^{2}}$$.
Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).
f is increasing on the intervals
f is decreasing on the intervals
The relative maxima of f occur at x =
The relative minima of f occur at x =
Notes: In the first two, your answer should either be a single interval, such as (0,1), a comma separated list of intervals, such as (-inf, 2), (3,4), or the word "none".
In the last two, your answer should be a comma separated list of x values or the word "none".

2021-03-12
$$f(x) = 4 - \frac{2}{x}+\frac{6}{x^{2}}$$
Take derivative and find out critical points to get maxima and minima
Apply power rule
$$f(x) = 4 - \frac{2}{x}+\frac{6}{x^{2}}$$
$$f(x) = 0 + \frac{2}{x^{2}}+\frac{12}{x^{3}}$$
$$f(x) = \frac{2}{x^{2}}+\frac{12}{x^{3}}$$
Set the derivative $$=0$$ and solve for x
LCD is $$x^{3}$$,
multiply each term by $$x^{3}$$
$$\frac{2}{x^{2}}-\frac{12}{x^{3}}=0$$
$$\frac{2}{x^{2}}(x^{3})-\frac{12}{x^{3}}(x^{3})=0(x^{3})$$
$$2x - 12 = 0$$
$$2x = 12$$
$$x = 6$$
$$x=0$$ makes denominator 0 and it is undefined
So $$x=0, x=6$$
Break the numbers line into three intervals using 0 and 6
Check each interval using derivative Let $$x=-1 , x=1, x=7$$
$$f'(x)=\frac{2}{x^{2}}-\frac{12}{x^{3}}$$
$$f'(-1)=\frac{2}{(1)^{2}}-\frac{12}{(1)^{3}}=-10$$
$$f'(1)=\frac{2}{(1)^{2}}-\frac{12}{(1)^{3}}=-10$$
$$f'(7)=\frac{2}{(7)^{2}}-\frac{12}{(7)^{3}}=\frac{2}{343}$$
Derivative is from positive in two intervals
So increasing intervals are$$(-\infty,0) U(6, \infty)$$
Decreasing interval $$(0, 6)$$
There is a break in graph of $$f(x) at x = 0$$
The derivative goes from negative to positive at $$x = 6$$
So relative minima at $$x = 6$$
There is no relative maxima

### Relevant Questions

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A. Let y=f(x) be the equation of C. Find f(x).
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Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
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$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
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The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
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At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
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$$\mu_1 - \mu_2$$.
(Round your answers to two decimal places.)
lower limit
upper limit
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Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
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