To find the values of $x$ for which the curve $y = x^3 - x^2 + 9$ has a tangent line parallel to the line $y = 5x + 1$, we need to understand that parallel lines have the same slope.

Solution:

1. Determine the slope of the given line $y = 5x + 1$:

   • The slope of $y = 5x + 1$ is $5$, since it’s the coefficient of $x$.

2. Find the derivative of the curve $y = x^3 - x^2 + 9$:

   • To find the slope of the tangent line to the curve, we differentiate $y = x^3 - x^2 + 9$ with respect to $x$: $\frac{dy}{dx} = 3x^2 - 2x$

3. Set the derivative equal to 5 to find the points where the tangent slope is 5: $3x^2 - 2x = 5$

   • Rearrange this equation to: $3x^2 - 2x - 5 = 0$

4. Solve the quadratic equation $3x^2 - 2x - 5 = 0$:

   • Using the quadratic formula: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3}$ $x = \frac{2 \pm \sqrt{4 + 60}}{6}$ $x = \frac{2 \pm \sqrt{64}}{6}$ $x = \frac{2 \pm 8}{6}$
     • This gives: $x = \frac{10}{6} = \frac{5}{3} \quad \text{and} \quad x = \frac{-6}{6} = -1$

5. Conclusion:

   • The values of $x$ where the curve has a tangent line parallel to $y = 5x + 1$ are $x = \frac{5}{3}$ and $x = -1$.

So, the correct answer is:

• (c) $x = \frac{5}{3}$ and $x = -1$.

Would you like further details on any step?

Here are some related questions to deepen understanding:

1. How do you determine the slope of a tangent line to a curve at a specific point?
2. What is the geometric meaning of setting the derivative equal to the slope of a given line?
3. How would the solution change if the line’s equation was different, such as $y = 3x + 4$?
4. Why is the quadratic formula useful in solving this type of problem?
5. How can you verify that the points $x = 5/3$ and $x = -1$ actually give a tangent parallel to the line?

Tip: When you encounter a curve with a tangent parallel to a line, the first step is always to match the slopes by using the derivative.