Question

Determine all values of x, (if any), at which the graph of the function has a horizontal tangent.
y(x)= (6x)/((x-9)^2)

Answer 1

Help please

Answer 2

Hey Becks c:

So uhhh how far did you get?
Looks like we need to apply quotient rule, yes?

Answer 3

\[\Large\rm y(x)=\frac{6x}{(x-9)^2}\]
\[\Large\rm y'(x)=\frac{\color{royalblue}{\left[6x\right]'}(x-9)^2-6x\color{royalblue}{\left[(x-9)^2\right]'}}{\left[(x-9)^2\right]^2}\]

Answer 4

So there is our quotient rule setup, yah? :d

Answer 5

yes, I got that far.

Answer 6

Taking our derivatives, setting the y' equal to zero gives us,\[\Large\rm 0=\frac{\color{orangered}{\left[6\right]}(x-9)^2-6x\color{orangered}{\left[2(x-9)\right]}}{\left[(x-9)^2\right]^2}\]

Answer 7

We want to look at the denominator and numerator `separately`.

Answer 8

Recall that our original function had a \(\Large\rm (x-9)^2\) in the denominator.
Which told us that x=9 is NOT in the domain of the function.

So it turns out that the denominator of our derivative function isn't giving us any horizontal tangents. (Since 9 isn't a value we can consider).
Actually there is another reason... critical values from the denominator usually correspond to vertical tangents, not horizontal.. blahhh let's not worry about that.

Answer 9

Set your numerator equal to zero and solve for x.

Answer 10

\[\Large\rm 0=6(x-9)^2-12x(x-9)\]

Answer 11

What do you think Becks? Too confusing? c:

Answer 12

Just a little bit. :p So the numerator doesn't have any horizontal tangents either?

Answer 13

It doesn't? :o
Hmm it probably does.

Got to expand everything out in order to solve for x.
Do you remember how to expand a square binomial?\[\Large\rm (x-9)^2=?\]

Answer 14

(x-9)(x-9)

Answer 15

Nooo silly, expand it out.
FOIL it out.

Answer 16

x^2-18x+81

Answer 17

\[\Large\rm 0=6\color{green}{(x-9)^2}-12x(x-9)\]\[\Large\rm 0=6\color{green}{(x^2-18x+81)}-12x(x-9)\]Mmmm ok good.

Answer 18

Distribute your 6,
Distribute your -12x,

and then combine like terms.
What are you left with?
Take a minute, try to get through those steps! c:

Answer 19

9=x

Answer 20

Well we already determined that x=9 isn't in the domain of our function.
You're missing a value though :O

If you simplified correctly, you should get:\[\Large\rm 0=486-6x^2\]Divide a 6 out,\[\Large\rm 0=81-x^2\]Which will factor to become,\[\Large\rm 0=(9-x)(9+x)\]

Answer 21

If you had done it the other way:\[\Large\rm \frac{486}{6}=x^2\]Then you need to remember that when you take the `square root of a square`, you need a plop a +/- symbol in front.

Answer 22

+/-9? so it would only be -9 because x= 9 is not in the domain of the function?

Answer 23

Yessss, good :)
x=-9 is our only solution for horizontal tangent line.

Answer 24

thank you!

Answer 25

yay team \c:/