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anonymous
 5 years ago
Application of Differentiation ( max and min values)
Help me through these steps to find max and min.... f(x)= x/x^2+1, [0,2]
anonymous
 5 years ago
Application of Differentiation ( max and min values) Help me through these steps to find max and min.... f(x)= x/x^2+1, [0,2]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The max/min will occur at an end point of your interval or at an interior point where \[f^{\prime}(x)=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is f(x) in this case?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it\[f(x)=\frac{x}{x^2+1}\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is difficult to tell by the way you wrote it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, I get that your are looking for absolute max/min on the interval [0,2]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay so first we need \[f^\prime(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you have it ? or do we need to figure it out?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f^\prime(x)=\frac{1x^2}{(x^2+1)^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes its 1(x^2+10 * x(2x+1) / (x^2+1)^2 right....i dont know how to write it out how you are

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can use the equation writer button or encode directly with Latex as I am, but its okay to use plain text as long as you use grouping symbols to include all of the numerator and denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your derivative does not look correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f^\prime(x)=\frac{(1)(x^2+1)x(2x)}{(x^2+1)^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i so knew that but rushing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{x^2+12x^2}{(x^2+1)^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then we can move forward

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since the denominator is the sum of positive numbers the derivative of this function is never undefined so we just need to find where it equals 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0setting the numerator equal to zero we get \[x=\pm 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02x^2 / x^2+1 is what i get next

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now note that \[x=1\] is not in the interval [0,2] so we do not need to consider it for this problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we just evaluate \[f(0),f(1),f(2)\] and compare

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(0)=0\]\[f(1)=\frac{1}{2}\]\[f(2)=\frac{2}{5}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so on the interval [0,2] our function has an absolute max of 1/2 and an absolute min of 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay i get the last part you did but after i get the x^2+12x^2 / (x^2+1)^2 i get confused on what you mean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the numerator of your derivative is 1x^2 after you "clean it up"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats the thing where i really need help my college algebra suck i dont know how to get that after i get 2x^2 / x^2+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay i can see that i just thought you would canceled the numerator x^2+1 with the denominator and the square would cancel
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