Question

find the equation of the tangent to the inverse of the function f(x)=x^2+8x+cos3x at (1,0)

Answer 1

i need help please.

Answer 2

Firstly find the equation of the tangent at that point, for the original function. Then find the inverse of that line (I think).

I will work it out and check that it is right.

Answer 3

okay

Answer 4

please try and help me work it out.

Answer 5

I think I've got it, do you have an answer from the textbook or do you not know the answer?

Answer 6

i do not know the answer.

Answer 7

\[y=x^2+8x+\cos(3x)\]
\[y'=2x+8-3\sin(3x)\]

Now we need to find the derivative at the point, but, since we want the tangent at the point (1,0) on the inverse function, we must inverse this point because we are going to use the derivative of the original function. We do this by swapping the coordinates.

(0,1)

Answer 8

Now, we find the value of the derivative at (0,1)

\[y'=2\times0+8-3\sin(3\times0)\]
\[y'=8\]

Therefore m=8 going through the point (0,1)
\[y-y_1=m(x-x_1)\]
\[y-1=8(x-0)\]
\[y=8x+1\]

Answer 9

Now the final step is to inverse the tangent.
\[y=8x+1\]
\[\frac{y-1}{8}=x\]

So the tangent of the inverse at (1,0) is:
\[y=\frac{1}{8}x-\frac{1}{8}\]

Make sense?

Answer 10

so You don't have to find the inverse of the original function?

Answer 11

looking through my textbook now i saw the answer to be
\[y =\frac{ 1 }{ 8 }\left( x -1 \right)\]

Answer 12

have understood now . you have got the right answer. I'm grateful

Answer 13

but what of if you are asked to find the eqn to the tangent of the function f(x)=x^2+8x+sinx at (0,9)

Answer 14

You take the inverse of the point (0,9); find the equation of the tangent for the original function at that point; inverse the equation for the tangent. It's exactly the same steps, give it a go and if you get stuck then ask for help :)

Answer 15

the question did not state the inverse of the function but just for the function. i still have to use the inverse?

Answer 16

Oh, sorry, I didn't see that. You just have to follow the steps without inversing the point and without inversing the tangent equation at the end:

Find derivative of function > find equation passing through the specified point

Answer 17

i have got y=9x+9
i'm i on track

Answer 18

Are you sure you copied the question correctly?
According to this there is no tangent at that point

http://www.wolframalpha.com/input/?i=y%3Dx%5E2%2B8x%2Bsinx+tangent+at+%280%2C9%29

Answer 19

that is what the question stated

Answer 20

I have to go now. Feel free to use that site as a way to check your answers, and, if you are still stuck then try bumping this question up to the top or making a new question and tagging some of the other helpers online.

There is always the chance that the question is written wrong too.

Bye-bye

Answer 21

thank you .

Answer 22

i wiil use what you taught me to try the inverse of the funtion, if i can get the answer.

Answer 23

i'm very grateful for your help.

Answer 24

The inverse will also not work, the only coordinate with a 0 in it that will work for the function you gave me is (0,0). You can try that as a practice and then check it using wolfram if you like.

If you liked my help then you can give me a medal by pressing the "best response" button on the right :)

Answer 25

okay, i have given you the medal

Answer 26

enjoy your day.