To find the tangent line to the graph of the function $f(x) = x - x^3$ at the point $(1, 0)$, follow these steps:

Step 1: Find the derivative of $f(x)$

The derivative of $f(x)$ will give us the slope of the tangent line at any point on the graph.

$f'(x) = \frac{d}{dx}(x - x^3) = 1 - 3x^2$

Step 2: Evaluate the derivative at $x = 1$

The slope of the tangent line at $x = 1$ is:

$f'(1) = 1 - 3(1)^2 = 1 - 3 = -2$

Step 3: Use the point-slope form of the equation of a line

The point-slope form of a line is given by:

$y - y_1 = m(x - x_1)$

Where $m$ is the slope and $(x_1, y_1)$ is the point of tangency, which in this case is $(1, 0)$.

Substituting the values:

$y - 0 = -2(x - 1)$

Simplifying:

$y = -2x + 2$

Final Answer

The equation of the tangent line to the graph of $f(x) = x - x^3$ at the point $(1, 0)$ is:

$y = -2x + 2$

Would you like more details or have any other questions?

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Here are five related questions you can explore:

1. How do you find the normal line to a curve at a given point?
2. What is the geometric meaning of the derivative at a point?
3. Can a curve have multiple tangent lines at a single point?
4. How do you find the second derivative of $f(x) = x - x^3$?
5. What is the concavity of the graph of $f(x) = x - x^3$ at $x = 1$?

Tip: When finding the equation of a tangent line, always remember to substitute both the slope and the point into the point-slope form to avoid mistakes.