Find an equation of the tangent line to the graph of the function f

In summary, the equation of the tangent line to the graph of the function f at the point (1, -1) is y3-x3=1.
  • #1
A_Munk3y
72
0

Homework Statement


Find an equation of the tangent line to the graph of the function f defined by the following equation at the indicated point.
(x - y - 1)3 = x; (1, -1)

The Attempt at a Solution


x3-y3=1
3y2(dy/dx)-3x2=0
3y2(dy/dx)=3x22
(dy/dx)=3y2/3x2
(dy/dx)=x2/y2
slope = (dy/dx) = 1

y-(-1)= 1(x-1)
y=x-2

thats wrong -.-
i tried making it dy/dx = x/y and still wrong.
where am i messing up? I'm pretty sure it has to do with the x at the end but i have no idea what to do with it!
We just learned derivatives so I'm still messing up with them.
 
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  • #2
What has x^3-y^3=1 got to do with the problem? Your equation is (x-y-1)^3=x. Find y' using implicit differentiation.
 
  • #3
Yea, i thought i was deriving it wrong :) Ok, let me go back and look at my notes to see how to do implicit differentiation.
Thanks
 
  • #4
... i can't figure out how to do the implicit differentiation on this. Could someone help me out with it?
 
  • #5
A_Munk3y said:
... i can't figure out how to do the implicit differentiation on this. Could someone help me out with it?

The derivative of u^3 is 3*u^2*u'. Put u=x-y-1. What do you get? Is that the part that's confusing you?
 
  • #6
I'm getting confused on the dx/dy part. I'm not even sure i understand how to do this right.
I thought i had to have them all cubed then get the derivative of x3-y3-13 but obviously that's not right since your first response was what did x3-y3= 1 have to do anything.

So is it 3x2*u and 3y2*u -1? and then i add the dx/dy part somewhere? Or am i just way off here :(
Sorry, we just learned this stuff today and it still hasn't really sunk in.
 
  • #7
Why would you implicitly differentiate? Since 3 is an odd power, you can solve for y without losing any information...
 
  • #8
Char. Limit said:
Why would you implicitly differentiate? Since 3 is an odd power, you can solve for y without losing any information...

Good point. If you aren't comfortable with implicit differentiation, try it that way. They both work.
 

1. What is the equation of a tangent line?

The equation of a tangent line is a linear equation that represents the slope of the curve at a specific point. It can be written in the form y = mx + b, where m is the slope of the tangent line and b is the y-intercept.

2. How do you find the equation of a tangent line?

To find the equation of a tangent line, you need to first find the slope of the curve at the point of interest. This can be done by taking the derivative of the function. Then, plug in the x-value of the point into the derivative to find the slope. Finally, use the point-slope form of a line to write the equation of the tangent line.

3. What information do you need to find the equation of a tangent line?

To find the equation of a tangent line, you will need the function f(x), the x-value of the point of interest, and the first derivative of the function.

4. Can there be more than one tangent line to a curve?

Yes, there can be more than one tangent line to a curve. This happens when the curve has a point of inflection or a point where the slope is undefined. In these cases, there can be multiple tangent lines at that point.

5. How does the equation of a tangent line relate to the graph of a function?

The equation of a tangent line represents the slope of the curve at a specific point. This slope is also visible on the graph of the function, as the tangent line touches the curve at that point and shows the direction in which the curve is changing.

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