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)

p=(x1, y1)

tangent line:
y-y1=m(x-x1)

normal line:
y-y1=(-1/m)(x-x1)

where m=slope=y'
(hint: solve for y. Then find y')

http://www.cliffsnotes.com/study_guide/Tangent-and-Normal-Lines.topicArticleId-39909,articleId-39887.html

Ignore the normal line formula. Your question doesn't ask for an equation of the normal line.

To find the equation of the tangent line to the graph of the function f at the given point, you will need to find the derivative of the function and evaluate it at the given point.

Step 1: Find the derivative of the function f.
To find the derivative, you need to differentiate both sides of the given equation with respect to x.

Differentiating both sides of the equation (x - y - 1)^3 = x, using the chain rule, we have:

3(x - y - 1)^2 * (1 - dy/dx) = 1

Now, isolate dy/dx by moving other terms to the opposite side:

3(x - y - 1)^2 * (1 - dy/dx) = 1
3(x - y - 1)^2 = 1 + 3dy/dx
dy/dx = (3(x - y - 1)^2 - 1)/3

Step 2: Evaluate the derivative at the given point (1, -1).
Substitute x = 1 and y = -1 into the derivative expression:

dy/dx = (3(1 - (-1) - 1)^2 - 1)/3
dy/dx = (3(1 + 1 - 1)^2 - 1)/3
dy/dx = (3(1)^2 - 1)/3
dy/dx = (3 - 1)/3
dy/dx = 2/3

Step 3: Use the slope-intercept form to determine the equation of the tangent line.
The equation of the tangent line can be written in the slope-intercept form as:

y - y1 = m(x - x1)

where (x1, y1) is the given point, and m is the slope of the tangent line.

Substituting the given point (1, -1) and the slope dy/dx = 2/3 into the equation, we have:

y - (-1) = (2/3)(x - 1)
y + 1 = (2/3)(x - 1)

Simplifying, we get the equation of the tangent line:

y + 1 = (2/3)x - 2/3
y = (2/3)x - 5/3

Therefore, the equation of the tangent line to the graph of the function f at the point (1, -1) is y = (2/3)x - 5/3.