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)
Why did the function f cross the road? To find the equation of the tangent line at (1, -1) of course! Let's get started.
First, let's differentiate the equation (x - y - 1)^3 = x with respect to x to find the slope of the tangent line. Using the chain rule, the derivative is:
3(x - y - 1)^2 * (1 - dy/dx) = 1
Now we can plug in the values for the point (1, -1) to find dy/dx:
3(1 - (-1) - 1)^2 * (1 - dy/dx) = 1
3(3)^2 * (1 - dy/dx) = 1
27(1 - dy/dx) = 1
27 - 27(dy/dx) = 1
27(dy/dx) = 26
dy/dx = 26/27
So the slope of the tangent line is 26/27. Now let's use the point-slope form of the equation of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Plugging in the values from the point (1, -1) and the slope 26/27:
y - (-1) = (26/27)(x - 1)
Simplifying this, we get:
y + 1 = (26/27)x - 26/27
And there you have it! The equation of the tangent line to the graph of the function f at the point (1, -1) is:
y = (26/27)x - 53/27
Now that's what I call a funny tangent line!
To find the equation of the tangent line to the graph of the function f at the point (1, -1), follow these steps:
Step 1: Differentiate both sides of the equation (x - y - 1)^3 = x with respect to x to find the derivative of f.
Let's simplify the equation first:
(x - y - 1)^3 = x
Expand the equation:
(x - y - 1)(x - y - 1)(x - y - 1) = x
(x^2 - 2xy + y^2 - 2x + 2y + 1)(x - y - 1) = x
(x^3 - 3x^2y - x^2 - xy^2 + 3xy + 2x + y^3 - y^2 - 2y - x + y + 1) = x
x^3 - 3x^2y - x^2 - xy^2 + 3xy + 2x + y^3 - y^2 - 2y - x + y + 1 = x
x^3 - 3x^2y - x^2 - xy^2 + 3xy + 2x + y^3 - y^2 - 2y - x + y + 1 - x = 0
x^3 - x^2 - xy^2 + 2xy + 2x + y^3 - y^2 - y + 1 = 0
Now, differentiate both sides with respect to x:
d/dx [x^3 - x^2 - xy^2 + 2xy + 2x + y^3 - y^2 - y + 1] = d/dx [0]
3x^2 - 2xy - y^2 + 2y + 2 = 0
Step 2: Substitute the coordinates of the point (1, -1) into the derivative equation to find the slope of the tangent line.
Substitute x = 1 and y = -1 into the equation:
3(1)^2 - 2(1)(-1) - (-1)^2 + 2(-1) + 2 = 0
3 - 2 + 1 - 2 + 2 = 0
2 = 0 (which is false)
This means that the point (1, -1) is not on the graph of the function. Please double-check the given point or equation.
If you have any other questions, feel free to ask!
To find the equation of the tangent line to the graph of the function at the indicated point, we need to find the slope of the tangent line and then use the point-slope form of a line to write the equation.
Step 1: Find the derivative of the function.
Differentiate both sides of the equation (x - y - 1)^3 = x with respect to x using the chain rule. The derivative of (x - y - 1)^3 is 3(x - y - 1)^2 * (1 - dy/dx), and the derivative of x is 1. So, we get:
3(x - y - 1)^2 * (1 - dy/dx) = 1
Step 2: Plug in the coordinates of the given point.
Since the point given is (1, -1), we substitute x = 1 and y = -1 into the equation from step 1:
3(1 - (-1) - 1)^2 * (1 - dy/dx) = 1
Simplifying:
3(1 + 1 - 1)^2 * (1 - dy/dx) = 1
3(1)^2 * (1 - dy/dx) = 1
3(1) * (1 - dy/dx) = 1
3 * (1 - dy/dx) = 1
3 - 3(dy/dx) = 1
Step 3: Solve for dy/dx.
Move the terms involving dy/dx to one side:
-3(dy/dx) = 1 - 3
-3(dy/dx) = -2
Divide both sides by -3:
dy/dx = -2/-3
dy/dx = 2/3
Step 4: Write the equation using point-slope form.
Now that we have the slope of the tangent line at the point (1, -1), we can use the point-slope form of a line which is y - y1 = m(x - x1).
Substitute the slope (2/3) and the coordinates of the given point (1, -1) into the equation:
y - (-1) = (2/3)(x - 1)
Simplify:
y + 1 = (2/3)(x - 1)
This is the equation of the tangent line to the graph of the function f at the point (1, -1).
I will assume your equation is
(x - y - 1)^3 = x
then 3(x-y-1)^2(1 - dy/dx) = 1
at the given point...
3(1+1-1)^2(1-dy/dx) = 1
3 - 3dy/dx = 1
dy/dx = 2/3
so now you have the slope m=2/3 and the point (1,-1)
Use the method you learned in grade 9 to find the equation of that line.