Question

Determine the values of x where the tangent line is horizontal for the function:
1/(x2−16)(x−7)
The value(s) of x where the tangent line to the graph of the function is horizontal is(are)_.

I know that the derivative of the f(x) is
f′(x)=−3x2+14x−16/(x−7)^2(x2−16)^2...
but I'm confused on where to do next

Answer 1

where to *go ><

Answer 2

Well, just deal with the numerator. Anything that would cause a 0 in the denominator would just be giving you a spot that is not differentiable. So just focus on the numerator and see if you can factor that to find your zeros and hence, the horizontal tangent lines.

Answer 3

There are convenient values, by the way : )

Answer 4

-16/3 and -2/3?

Answer 5

wait no that's not right..

Answer 6

Think you can correct the error?

Answer 7

i'm going over my work but i'm not sure what i'm doing wrong.

Answer 8

WAIT i think i got it lol

Answer 9

So what do ya got? : )

Answer 10

Just a question @jbeloy17.... have you tried graphing this first? I know you have to do it algebraically as well, but try graphing it first to gain a visual perspective.

Answer 11

i got 2 and -8/9 this time but that's not right..

Answer 12

\[-3x^{2}+14x-16 = -(3x^{2}-14x+16) = -(3x^{2}-6x-8x+16)\]Now factoring by grouping:
\[-((3x^{2}-6x)+(-8x+16))\]
\[-(3x(x-2)-8(x-2))\]
\[-(3x-8)(x-2)\]

Answer 13

@Hero i graphed it already but i need the exact values

Answer 14

okay.. i kinda get it.

Answer 15

Is something still unclear, though?

Answer 16

no i get it. i tried to do borrow and payback, or just minusing the 16 from the beginning. but yeah i get it.

Answer 17

now i just set -(3x-8)(x-2)=0?

Answer 18

Each factor, yes.
3x-8 = 0
x-2 = 0

Answer 19

2 and 8/3

Answer 20

Bingo

Answer 21

okay!... but i put it into my homework online, and it says that it's not the right answer

Answer 22

@psymon idk what i'm doing wrong!

Answer 23

Im trying to check it. I personally do not see an inssue, but I could easily be missing something.

Answer 24

okay. thank you for your time!

Answer 25

anything @psymon?

Answer 26

Not really. This is kinda random, but try
\[\frac{ -7 \pm \sqrt{97} }{ 3 }\]

Answer 27

@jbeloy17 !!!!!You just made a sign mistake on the derivative x_x

Answer 28

what?! ><

Answer 29

.9496192673 and -5.616285934 ???

Answer 30

\[\frac{ 1 }{ (x^{2}-16)(x-7) }= (x^{2}-16)^{-1}(x-7)^{-1}\]
\[-1(x^{2}-16)^{-2}(2x)(x-7)^{-1} + (x^{2}-16)^{-1}(-1)(x-7)^{-2}\]
\[\frac{ -2x }{ (x^{2}-16)^{2}(x-7) }-\frac{ 1 }{ (x^{2}-16)(x-7)^{2} }\]
\[\frac{ -2x(x-7)-(x^{2}-16) }{ (x^{2}-16)^{2}(x-7)^{2} }= \frac{ -2x^{2}+14x-x^{2}+16 }{ (x^{2}-16)^{2}(x-7)^{2} }\]
\[\frac{ -3x^{2}+14x+16 }{ (x^{2}-16)^{2}(x-7)^{2} }\]\[\frac{ -14 \pm \sqrt{14^{2} - 4(-3)(16)} }{ -6 } = \frac{ -14 \pm 2\sqrt{97} }{ -6 }= \frac{ -7 \pm \sqrt{97} }{ 3 }\]

Answer 31

Leave it in radical form, dont turn it into decimals unless it instructsyou to.

Answer 32

it says that i have to find the values

Answer 33

Well, those are the values. Theyre the EXACT values. Turning it into a decimal just turns the answer into an approximation.

Answer 34

i put the answers in and it says that it's incorrect

Answer 35

Then try the decimals I suppose. But thats correct.

Answer 36

Unless you see an error in my math above :/

Answer 37

thanks for everything @psymon really it means a lot!!

Answer 38

Did inputting decimals work for you @jbeloy17 ?

Answer 39

Yeah, np. Did what I could, lol.

Answer 40

no it didn't

Answer 41

nope

Answer 42

lol, okay

Answer 43

What was your suspicion, hero?

Answer 44

Just trying something, that's all. Pay me no attention.

Answer 45

Lol, alrighty then.

Answer 46

Is that the exact wording of the question? @jbeloy17

Answer 47

Determine the values of where the tangent line is horizontal for the function:

jbeloy17
Determine the values of x where the tangent line is horizontal for the function:
1/(x^2−16)(x−7)
The value(s) of where the tangent line to the graph of the function is horizontal is(are

Answer 48

sorry copied it twice

Answer 49

Alright, just making sure in case there could have been a misinterpreting of the question.