Question

given the function g(x)=x^2/(x^2+1) find the following:
a) State the intervals of increase and decrease in g.
b) State the intervals of upward and downward concavity in g.
Show the calculations that lead to your conclusions.

Answer 1

for a you need to find the critical numbers

Answer 2

first we need the derivative to do that

Answer 3

\[f'(x)=\frac{(x^2)'(x^2+1)-x^2(x^2+1)'}{(x^2+1)^2}=\frac{2x(x^2+1)-x^2(2x+0)}{(x^2+1)^2}\]

Answer 4

\[f'(x)=\frac{2x(x^2+1)-2x^3}{(x^2+1)^2}=\frac{2x(x^2+1-x^2)}{(x^2+1)^2}=\frac{2x}{(x^2+1)^2}\]

Answer 5

is that what you got?

Answer 6

for f'?

Answer 7

now x^2+1 is never 0 so we don't have to worry with f' being undefined
but 2x=0 when x=0
so x=0 is a critical number

Answer 8

yes that is what i got so far

Answer 9

ok great

Answer 10

But then from there i take the derivative of the last equation you got right?

Answer 11

so test the areas around x=0 with f' to see if the function is decreasing there or increasing

Answer 12

lets do b after we have completely done a

Answer 13

unless you totally understand a

Answer 14

so then lets plug in like -1 & 1 in order to test it?

Answer 15

ok those are good numbers

Answer 16

so we plug in -1 &1 into what equation though?

Answer 17

its obvious that f will be decreasing before x=0 since f'<0
and f will be increasing after x=0 since f'>0

Answer 18

f'

Answer 19

f' tells if a function is increasing or decreasing

Answer 20

so you use f' to test for that

Answer 21

oh okay now i get it! thanks so now we work on b?

Answer 22

ok by the way i called your function f instead of g
did you notice that?

Answer 23

just replace my f's with g's okay?

Answer 24

yes i did notice that.

Answer 25

no worries i will fix that.

Answer 26

yes we do.

Answer 27

so anyways we found f' which tells us increasing and decreasing
now we need find f'' which tells about f's concavity

Answer 28

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

Answer 29

\[f''(x)=\frac{2(x^2+1)^2-2x \cdot 2(x^2+1)(2x)}{(x^2+1)^4}\]

Answer 30

is that okay so far?

Answer 31

do you understand what i did?

Answer 32

yes i understand what you did so far

Answer 33

\[f''(x)=\frac{2(x^2+1)[(x^2+1)-2x (2x)]}{(x^2+1)^4}=\frac{2(x^2+1)[(x^2+1)-4x^2]}{(x^2+1)^4}\]

Answer 34

let me know if you see a mistake

Answer 35

\[f''(x)=\frac{2(x^2+1)(-3x^2+1)}{(x^2+1)^4}\]

Answer 36

now f'' is 0 when -3x^2+1=0 because x^2+1 is not zero for any real value x

Answer 37

\[3x^2=1 => x^2=\frac{1}{3} => x=\pm \sqrt{\frac{1}{3}}\]

Answer 38

now to see if these are inflection numbers we need to test the the concavity around these numbers by choosing test numbers around these numbers and plug them into f''

Answer 39

so we know 0 is in between those numbers
and we know -5 is before -sqrt(1/3) and 5 is after sqrt(1/3)

so we can use these to test the intervals

Answer 40

\[f''(-5)=\frac{2((-5)^2+1)(-3(-5)^2+1)}{((-5)^2+1)^4}=\frac{(+)(-)}{(+)}=(-)\]
so on the interval (-inf,-sqrt(1/3)) the function is concave down

Answer 41

\[f''(0)=\frac{2(+)(+)}{(+)}=(+) \]
=> (-sqrt(1/3),sqrt(1/3) is concave up

Answer 42

i will leave the last interval for you

Answer 43

the last interval we are using is 5 right?

Answer 44

you don't have to use 5 that is just the number i chose after sqrt(1/3)

Answer 45

okay ill just use 5. so 2((5)^2+1)(-3(5)^2+1)/ ((5)^2+1)^4=(+)(-)/(+) so it is concave downward? right?

Answer 46

right :)

Answer 47

yay! (: do you think you could possible help me with just one more question for calculus?

Answer 48

post a new thread and i might get to it okay?
:)

Answer 49

okay thanks! have a great day (: