To find the value of $k$ such that the line $y = 10x + 1$ is tangent to the graph of the function $f(x) = kx^3$, we need to ensure two conditions are met:

1. The line and the curve must intersect at some point $x = a$.
2. The derivative of the curve at $x = a$ must be equal to the slope of the line (10 in this case).

Step 1: Find the Intersection

Setting the equations equal to find the intersection point $a$: $k a^3 = 10a + 1$

Step 2: Set the Derivative Equal to the Slope of the Line

The derivative of $f(x) = kx^3$ is: $f'(x) = 3kx^2$ At $x = a$, this must equal the slope of the tangent line: $3ka^2 = 10$

Now, we solve these equations simultaneously to find $k$ and $a$.

Let's proceed with the calculations.The value of $k$ such that the line $y = 10x + 1$ is tangent to the graph of the function $f(x) = kx^3$ is $k = \frac{4000}{27}$. The point of tangency occurs at $x = -\frac{3}{20}$.

Would you like more details on any part of the solution or have any other questions? Here are some related topics you might find interesting:

1. How to find the point of intersection of the line and the curve.
2. What would happen if the slope of the tangent line were different?
3. How does the value of $k$ influence the shape of the curve $f(x) = kx^3$?
4. Deriving the general formula for the tangent line to a cubic function.
5. Practical applications of finding tangent lines to curves in real-world problems.

Tip: Understanding the relationship between a function and its derivatives can provide valuable insights into the behavior of the function, such as determining where it has maxima, minima, or points of inflection.