Question

determine the poits at which the graph of the function has a horizontal tangent: f(x)=x^2/x^2+1

Answer 1

what is f ' (x) ?

Answer 2

not sure...

Answer 3

you have to use the quotient rule

Answer 4

would i have to find it with the quotient rule ?

Answer 5

ooh okay

Answer 6

yes

Answer 7

\[2x(x^2+1)-(x^2)(2x+1) / (x^2 +1)^2\]
then i cancel out the top and bottom (x^2+1) ?

Answer 8

you can't do that cancellation

Answer 9

also, you should have 2x(x^2+1) - (x^2)*2 in the numerator

Answer 10

that simplifies to

2x*x^2 + 2x*1 - x^2*2

2x^3 + 2x - 2x^3

2x

Answer 11

after everything is simplified, you are left with

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

Answer 12

it is in the numerator :o
the denominator is the (x^2+1)^2 ... i typed it wrong ?

Answer 13

you had 2x+1 when it should be 2x

sorry I meant to say 2x and not just 2

Answer 14

ooh okay i see !

Answer 15

once you have \[\Large f^{\prime}(x) = \frac{2x}{(x^2+1)^2}\], you replace f ' (x) with 0

then solve for x

Answer 16

so it'll be one ?

Answer 17

nope

Answer 18

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

\[\Large 0 = \frac{2x}{(x^2+1)^2}\]

solve for x

Answer 19

0 ?

Answer 20

yes x = 0 is a solution, so the horizontal tangent is at (0,0)

Answer 21

you find y = 0 by plugging x = 0 back into f(x)

Answer 22

i struggle so much with ap calculus ;( do you have any advice that maybe i can take ??

Answer 23

well the key when you struggle with any material is to figure out what you know well already

Answer 24

then try to connect the new stuff to the old stuff you know

Answer 25

from my experience, calculus has a lot of algebra in it (example: simplifying complicated expressions). So if you struggle with topics like that, then I recommend you practice more with it.

As for the complicated derivative formulas, you can come up with clever ways to memorize them. Like with the quotient rule, you can use this trick

http://www.howtoace.com/HTACFiles/node21.html

or I like to think of it as a slightly modified form of the product rule (numerator' - denominator'). However works for you is fine

Answer 26

okay so whenever i am finding the tangent of a line i need to use the quotient rule ?

Answer 27

no, you use the quotient rule when deriving functions of the form

f(x) = g(x)/h(x)

Answer 28

you're given f(x)=x^2/x^2+1

so g(x) = x^2 and h(x) = x^2 + 1

Answer 29

ooh okay okay. and you always need to find the derivative in order to find the tangent ?

Answer 30

when they say "find the slope of the tangent line", you have to find f ' (x)

aka you have to differentiate or derive f(x) to get f ' (x)

Answer 31

the derivative f ' (x) is used to find the slope of the tangent line

Answer 32

yes that is correct

Answer 33

okay so whenever i need to find the tangent of a line i first have to find the derivative by using the quotient rule correct ?

Answer 34

"okay so whenever i need to find the tangent of a line i first have to find the derivative" correct

"... by using the quotient rule" only if you're dealing with a fraction of two functions

Answer 35

if you had f(x) = x^2

you just derive using the power rule (no need to use the quotient rule)

Answer 36

and if it isnt i use the power rule

Answer 37

okay got it

Answer 38

there are other rules as well. not just these two rules

Answer 39

the power rule is for power functions

a power function is in the form y = a*x^b

'a' and 'b' are constants

Answer 40

so a and b would become 0 ?

Answer 41

well in the case of y = x^2, which is really y = 1*x^2, a = 1 and b = 2