We are tasked with finding the value of $k$ such that the line $y = 7x + 6$ is tangent to the graph of the function $f(x) = kx^3$.

Steps:

1. The condition for tangency:
   For a line to be tangent to a curve at some point $x = a$, two conditions must hold:

   • The function values must be equal: $f(a) = 7a + 6$.
   • The slopes (derivatives) must be equal: $f'(a) = 7$.

2. Find the derivative of $f(x)$:
   The derivative of $f(x) = kx^3$ is: $f'(x) = 3kx^2$ At $x = a$, the slope of the function is: $f'(a) = 3ka^2$

3. Set the slopes equal:
   Since the slope of the tangent line is 7, we have: $3ka^2 = 7$ Solving for $k$, we get: $k = \frac{7}{3a^2}$

4. Set the function values equal:
   The function values at $x = a$ must also be equal, so: $f(a) = k a^3 = 7a + 6$ Substituting $k = \frac{7}{3a^2}$ into the equation: $\frac{7}{3a^2} \cdot a^3 = 7a + 6$ Simplifying: $\frac{7a}{3} = 7a + 6$ Multiply through by 3 to eliminate the fraction: $7a = 21a + 18$ Rearranging the terms: $-14a = 18$ Solving for $a$: $a = -\frac{9}{7}$

5. Substitute $a$ into the equation for $k$:
   Now that we have $a = -\frac{9}{7}$, substitute it into $k = \frac{7}{3a^2}$: $k = \frac{7}{3 \left( -\frac{9}{7} \right)^2} = \frac{7}{3 \cdot \frac{81}{49}} = \frac{7 \times 49}{243} = \frac{343}{243} = \frac{49}{35} = \frac{7}{5}$

Thus, the value of $k$ is $\frac{7}{5}$.

Would you like more details on any part of this explanation?

Here are five follow-up questions:

1. How do you find the derivative of a cubic function like $f(x) = kx^3$?
2. What is the geometrical interpretation of a line being tangent to a curve?
3. How would the problem change if the function were a higher degree polynomial?
4. What if the line had a different slope, say $y = 8x + 6$? How would the solution change?
5. How can we check if our solution is correct by plotting?

Tip: Always double-check your algebraic manipulations, especially when dealing with fractions and powers.