Question

determine the values of x where the tangent line is horizontal for the function
f(x)=1/((x^2-1)(x-4))

Answer 1

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

Answer 2

@hilbertboy96

Answer 3

ah these are always fun. Well tell me if you know when are tangent lines to a function horizontal? There is a special term we call them

Answer 4

i forgot the name, but i know that its when the slope/derivative is =0 right?

Answer 5

yes they are called critical points and that is when f'(x)=0

Answer 6

so we need to find f'(x) and the simpilest way to do this will most likely be to actually expand that bottom then do a composition rule

Answer 7

so the bottom would be \[x^3-4x^2-x+4\]

Answer 8

correct and to make it easier we can rewrite the function as

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

right?

Answer 9

yea

Answer 10

now we can take the derivative which as i said is a composition function so we apply the extended power rule

\[f'(x)=-(x^3-4x^2-x+4)^{-2}(3x^2-8x-1)\]

rewrite it and we get

\[\frac{ -3x^2+8x+1 }{(x^3-4x^2-x+4)^{2} }\]

and to make f'(x) = 0 we set the top equal to zero even though critical points are when f'(x) = 0 or does not exist but the do not exist spots are just vertical asymptotes

so

\[0=-3x^2+8x-1\]

Solve and you get your x-values

Answer 11

find them yet?

Answer 12

i got\[\frac{ -(\sqrt{13} -4)}{ 3 } and \frac{ \sqrt{13}+4 }{3 }\]

Answer 13

and it says its wrong :/

Answer 14

hmm rechecking work

Answer 15

wait did you go off of my 0=-3x^2+8x-1?

Answer 16

yea

Answer 17

the -1 should be +1

Answer 18

sorry lol

Answer 19

oh okay, let me change it

Answer 20

lol its okay

Answer 21

got it :), thanks, it came out to be
\[x=\frac{ -(\sqrt{19}-4) }{3 } and \frac{ \sqrt{19} +4}{ 3 }\]