Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Determine an equation of the line tangent to the graph of y= ((x^2-1)^2)/(x^3-6x-1) at the point (3,8)

Calculus1
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[y=\frac{(x^2-1)^2}{x^3-6x-1}\] Have you found y' yet?
Yes, I have but I am getting the final answer wrong, and I am not sure where I am making a mistake. It is the first time I can't figure out what I am doing wrong...
So you have \[((x^2-1)^2)'=2(x^2-1) \cdot 2x \] and \[(x^3-6x-1)'=3x^2-6\] ?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

So by quotient rule, we should have \[y'=\frac{2(x^2-1) \cdot 2x (x^3-6x-1)-(3x^2-6)(x^2-1)^2}{(x^3-6x-1)^2}\]
What did you get for the slope? I got a pretty little integer.
Say something so I can know you are still there. Like did you simplify that above and then plug in? I'm getting the slope of the tangent at x=3 is -9.
sorry. The answer is y=-9x+35
I got some bizzare answers that don't make much sense... However, for y' i got \[((x^2-1)(x^4-15x^2-4x-6))\div(x^3-6x-1)^2\]
myininaya how did you get -9 and 3?
Your question said to find the tangent line at x=3. I plugged x=3 into f'.
I did that and still did something wrong. Anyways, thanks for your help.
You y' is right... I don't understand why you aren't getting -9 when pluggin in 3.... \[\frac{(3^2-1)(3^4-15(3)^2-4(3)-6)}{(3^3-6(3)-1)^2}\] \[=\frac{(9-1)(81-135-12-6)}{(27-18-1)^2}=\frac{(8)(81-153)}{(27-19)^2}\] \[=\frac{(8)(-72)}{(8)^2}=\frac{8(-72)}{(8)(8)}=\frac{\cancel{8}(-72)}{\cancel{(8)}(8)}=\frac{-72}{8}=-9\]
You should be getting -9 as the slope of your line... So I don't know where you are going wrong in your calculation.
Hah, I found my mistake! I kept calculating 3^2 instead of 3^3 in the denominator...

Not the answer you are looking for?

Search for more explanations.

Ask your own question