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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
asked 2021-03-11
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".

Answers (1)

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

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