Let's break down the problem from the image.

Given Curve:

$f(x) = x - x^3$

(a) Find the slope of the tangent line to the curve at the point $(1, 0)$.

The slope of the tangent line at any point on the curve is given by the derivative of the function $f(x)$ evaluated at that point. So, let's start by finding the derivative of $f(x)$.

1. $f(x) = x - x^3$
2. The derivative is: $f'(x) = \frac{d}{dx}(x - x^3) = 1 - 3x^2$

Now, evaluate the derivative at $x = 1$: $f'(1) = 1 - 3(1)^2 = 1 - 3 = -2$ So, the slope of the tangent line at $(1, 0)$ is -2.

(b) Find the equation of the tangent line in part (a).

The equation of the tangent line is given by the point-slope form: $y - y_1 = m(x - x_1)$ Where:

• $m$ is the slope (found in part (a) as -2),
• $(x_1, y_1) = (1, 0)$ is the point of tangency.

Substitute the values: $y - 0 = -2(x - 1)$ Simplifying: $y = -2x + 2$ So, the equation of the tangent line is $y = -2x + 2$.

(c) Graph the curve and the tangent line.

From the image, the blue curve represents the function $f(x) = x - x^3$, and the red line represents the tangent line at $(1, 0)$, which is $y = -2x + 2$.

Would you like further details or explanations for any of these parts?

Here are 5 related questions you might explore:

1. How do you find the second derivative of $f(x) = x - x^3$?
2. What is the geometric interpretation of the second derivative at $(1, 0)$?
3. How does the behavior of the function change as $x \to \infty$ and $x \to -\infty$?
4. What are the critical points of $f(x) = x - x^3$, and what do they represent?
5. How do you find the concavity of the curve at different points?

Tip: Always verify the slope at the point of tangency by double-checking your derivative calculations!