## 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]

1. anonymous

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

2. anonymous

what is f(x) in this case?

3. anonymous

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

4. anonymous

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

5. anonymous

yes but also [0,2]

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. 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

11. 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

12. anonymous

your derivative does not look correct

13. anonymous

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

14. anonymous

i so knew that but rushing

15. anonymous

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

16. anonymous

okay, cool

17. anonymous

then we can move forward

18. 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

19. anonymous

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

20. NY,NY

habachuchu

21. anonymous

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

22. anonymous

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

23. anonymous

now we just evaluate $f(0),f(1),f(2)$ and compare

24. anonymous

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

25. anonymous

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

26. 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

27. anonymous

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

28. 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

29. anonymous

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

30. 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