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

- katieb

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- anonymous

The max/min will occur at an end point of your interval or at an interior point where \[f^{\prime}(x)=0\]

- anonymous

what is f(x) in this case?

- anonymous

is it\[f(x)=\frac{x}{x^2+1}\]?

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- anonymous

it is difficult to tell by the way you wrote it...

- anonymous

yes but also [0,2]

- anonymous

yes, I get that your are looking for absolute max/min on the interval [0,2]

- anonymous

okay so first we need \[f^\prime(x)\]

- anonymous

do you have it ? or do we need to figure it out?

- anonymous

\[f^\prime(x)=\frac{1-x^2}{(x^2+1)^2}\]

- anonymous

yes 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

you 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

your derivative does not look correct

- anonymous

\[f^\prime(x)=\frac{(1)(x^2+1)-x(2x)}{(x^2+1)^2}\]

- anonymous

i so knew that but rushing

- anonymous

\[=\frac{x^2+1-2x^2}{(x^2+1)^2}\]

- anonymous

okay, cool

- anonymous

then we can move forward

- anonymous

since 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

setting the numerator equal to zero we get \[x=\pm 1\]

- NY,NY

habachuchu

- anonymous

2x^2 / x^2+1 is what i get next

- anonymous

now note that \[x=-1\] is not in the interval [0,2] so we do not need to consider it for this problem

- anonymous

now we just evaluate \[f(0),f(1),f(2)\] and compare

- anonymous

\[f(0)=0\]\[f(1)=\frac{1}{2}\]\[f(2)=\frac{2}{5}\]

- anonymous

so on the interval [0,2] our function has an absolute max of 1/2 and an absolute min of 0

- anonymous

okay i get the last part you did but after i get the x^2+1-2x^2 / (x^2+1)^2 i get confused on what you mean

- anonymous

the numerator of your derivative is 1-x^2 after you "clean it up"

- anonymous

thats 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

x^2+1-2x^2=1-x^2

- anonymous

okay 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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