Let f(x)=4−2x+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".

Question

Let f(x)=4−2x+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 &quot;none&quot;.  In the last two, your answer should be a comma separated list of x values or the word &quot;none&quot;.

Talisha · Accepted Answer

f(x)=4−2x+6x2Take derivative and find out critical points to get maxima and minimaApply power rulef(x)=4−2x+6x2f‘(x)=0+2x2+12x3f‘(x)=2x2+12x3Set the derivative =0 and solve for xLCD is x3, multiply each term by x32x2−12x3=02x2(x3)−12x3(x3)=0(x3)2x−12=02x=12x=6x=0 makes denominator 0 and it is undefinedSo x=0,x=6Break the numbers line into three intervals using 0 and 6Check each interval using derivativeLet x=−1,x=1,x=7f′(x)=2x2−12x3f′(−1)=2(1)2−12(1)3=−10f′(1)=2(1)2−12(1)3=−10f′(7)=2(7)2−12(7)3=2343Derivative is from positive in two intervalsSo increasing intervals are(−∞,0)U(6,∞)Decreasing interval (0,6)There is a break in graph of f(x)atx=0The derivative goes from negative to positive at x=6So relative minima at x=6There is no relative maxima

Let f(x)=4−2x+6x2. Find the open intervals on which f is increasing (decreasing). Then determine the...

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