anonymous
  • anonymous
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]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
The max/min will occur at an end point of your interval or at an interior point where \[f^{\prime}(x)=0\]
anonymous
  • anonymous
what is f(x) in this case?
anonymous
  • anonymous
is it\[f(x)=\frac{x}{x^2+1}\]?

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anonymous
  • anonymous
it is difficult to tell by the way you wrote it...
anonymous
  • anonymous
yes but also [0,2]
anonymous
  • anonymous
yes, I get that your are looking for absolute max/min on the interval [0,2]
anonymous
  • anonymous
okay so first we need \[f^\prime(x)\]
anonymous
  • anonymous
do you have it ? or do we need to figure it out?
anonymous
  • anonymous
\[f^\prime(x)=\frac{1-x^2}{(x^2+1)^2}\]
anonymous
  • 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
  • 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
  • anonymous
your derivative does not look correct
anonymous
  • anonymous
\[f^\prime(x)=\frac{(1)(x^2+1)-x(2x)}{(x^2+1)^2}\]
anonymous
  • anonymous
i so knew that but rushing
anonymous
  • anonymous
\[=\frac{x^2+1-2x^2}{(x^2+1)^2}\]
anonymous
  • anonymous
okay, cool
anonymous
  • anonymous
then we can move forward
anonymous
  • 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
  • anonymous
setting the numerator equal to zero we get \[x=\pm 1\]
NY,NY
  • NY,NY
habachuchu
anonymous
  • anonymous
2x^2 / x^2+1 is what i get next
anonymous
  • 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
  • anonymous
now we just evaluate \[f(0),f(1),f(2)\] and compare
anonymous
  • anonymous
\[f(0)=0\]\[f(1)=\frac{1}{2}\]\[f(2)=\frac{2}{5}\]
anonymous
  • anonymous
so on the interval [0,2] our function has an absolute max of 1/2 and an absolute min of 0
anonymous
  • 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
  • anonymous
the numerator of your derivative is 1-x^2 after you "clean it up"
anonymous
  • 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
  • anonymous
x^2+1-2x^2=1-x^2
anonymous
  • 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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