The slope of the tangent line to the graph y= x^3/3-x at the point (1, -2/3)

So I plugged this into y= and I found the (1,-2/3) point but I don't know what to do next, can you help?

The answer is 0

Sure, I'd be happy to help! Finding the slope of the tangent line at a given point involves calculating the derivative of the function and plugging in the x-value of the point. Let's start by finding the derivative of the function:

y = (x^3/3) - x

To find the derivative, we'll use the power rule of differentiation. The power rule states that if we have a term of the form x^n, the derivative is nx^(n-1). Applying this rule to each term, we get:

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

Now that we have the derivative, we can calculate the slope of the tangent line at the point (1, -2/3) by plugging in x = 1:

slope = (1^2 - 1)
slope = (1 - 1)
slope = 0

So, the slope of the tangent line is 0. Keep in mind that the slope of zero indicates a horizontal line.

Yes, I can help you with that. To find the slope of the tangent line at a particular point on a curve, you need to take the derivative of the function and then evaluate it at that point.

Step 1: Find the derivative of the function y = x^3/3 - x.
To find the derivative, you need to apply the power rule and the constant rule. The power rule states that the derivative of x^n is n*x^(n-1). Applying this to our function, we get:

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

Step 2: Evaluate the derivative at the point (1, -2/3).
To find the slope of the tangent line at the point (1, -2/3), substitute x = 1 into the derivative equation:

slope = (1)^2 - 1
= 1 - 1
= 0

Therefore, the slope of the tangent line to the graph at the point (1, -2/3) is 0.

Yes, I can help you with that. To find the slope of the tangent line to the graph at a given point, you need to find the derivative of the function and then evaluate it at the given x-coordinate.

Step 1: Find the derivative of the given function y = x^3/3 - x. In this case, we can use the power rule, which states that the derivative of x^n is nx^(n-1). Applying the power rule, the derivative of x^3/3 is (1/3) * 3 * x^(3/3 - 1), which simplifies to x^(2/3).

Step 2: Evaluate the derivative at the x-coordinate of the given point, which is 1. Plug in x=1 into the derivative we found in step 1: x^(2/3). Thus, (1)^(2/3) = 1^(2/3) = 1.

Step 3: The resulting value is the slope of the tangent line at that point. In this case, the slope of the tangent line at (1, -2/3) is 1.

Therefore, the slope of the tangent line to the graph y = x^3/3 - x at the point (1, -2/3) is 1.

the tangent line has the same slope as the curve at the point of tangency. The slope at any x is

y' = x^2-1
So, y'(1) = 0

Now you have a point and a slope, so the line is

y + 2/3 = 0(x-1)
or,
y = -2/3

See the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3D+x^3%2F3-x%2C+y%3D-2%2F3