ruigE

2021-03-11

Let $f\left(x\right)=4-\frac{2}{x}+\frac{6}{{x}^{2}}$.
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".

Talisha

$f\left(x\right)=4-\frac{2}{x}+\frac{6}{{x}^{2}}$
Take derivative and find out critical points to get maxima and minima
Apply power rule
$f\left(x\right)=4-\frac{2}{x}+\frac{6}{{x}^{2}}$
$f‘\left(x\right)=0+\frac{2}{{x}^{2}}+\frac{12}{{x}^{3}}$
$f‘\left(x\right)=\frac{2}{{x}^{2}}+\frac{12}{{x}^{3}}$
Set the derivative $=0$ and solve for x
LCD is ${x}^{3}$,
multiply each term by ${x}^{3}$
$\frac{2}{{x}^{2}}-\frac{12}{{x}^{3}}=0$
$\frac{2}{{x}^{2}}\left({x}^{3}\right)-\frac{12}{{x}^{3}}\left({x}^{3}\right)=0\left({x}^{3}\right)$
$2x-12=0$
$2x=12$
$x=6$
$x=0$ 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 derivativeLet $x=-1,x=1,x=7$
${f}^{\prime }\left(x\right)=\frac{2}{{x}^{2}}-\frac{12}{{x}^{3}}$
${f}^{\prime }\left(-1\right)=\frac{2}{\left(1{\right)}^{2}}-\frac{12}{\left(1{\right)}^{3}}=-10$
${f}^{\prime }\left(1\right)=\frac{2}{\left(1{\right)}^{2}}-\frac{12}{\left(1{\right)}^{3}}=-10$
${f}^{\prime }\left(7\right)=\frac{2}{\left(7{\right)}^{2}}-\frac{12}{\left(7{\right)}^{3}}=\frac{2}{343}$
Derivative is from positive in two intervals
So increasing intervals are$\left(-\mathrm{\infty },0\right)U\left(6,\mathrm{\infty }\right)$
Decreasing interval $\left(0,6\right)$
There is a break in graph of $f\left(x\right)atx=0$
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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