Find the tangent line to the graph of f(x)=x-x^3 at the point (1,0)

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.

Learn how to find the tangent line to the graph of f(x) = x - x^3 at the point (1, 0). This calculus problem covers key concepts such as derivatives and the point-slope form. Suitable for high school students in Grades 10-12.

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