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

ruigE 2021-03-11 Answered
Let f(x)=42x+6x2.
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".
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Talisha
Answered 2021-03-12 Author has 93 answers
f(x)=42x+6x2
Take derivative and find out critical points to get maxima and minima
Apply power rule
f(x)=42x+6x2
f(x)=0+2x2+12x3
f(x)=2x2+12x3
Set the derivative =0 and solve for x
LCD is x3,
multiply each term by x3
2x212x3=0
2x2(x3)12x3(x3)=0(x3)
2x12=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(x)=2x212x3
f(1)=2(1)212(1)3=10
f(1)=2(1)212(1)3=10
f(7)=2(7)212(7)3=2343
Derivative is from positive in two intervals
So increasing intervals are(,0)U(6,)
Decreasing interval (0,6)
There is a break in graph of f(x)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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