We are tasked with finding the equation of the line tangent to the graph of $f(x) = \frac{6x}{x + 7}$ at the point $(1, 0.75)$.

Steps to solve:

1. Verify the point: First, we check that the point $(1, 0.75)$ is indeed on the curve.

   $f(1) = \frac{6(1)}{1 + 7} = \frac{6}{8} = 0.75$

   Since $f(1) = 0.75$, the point $(1, 0.75)$ is correct.

2. Find the derivative $f'(x)$: The derivative of $f(x) = \frac{6x}{x + 7}$ will give us the slope of the tangent line at any point.

   We can apply the quotient rule for derivatives, which states: $\left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$ where $u(x) = 6x$ and $v(x) = x + 7$.

   • $u'(x) = 6$
   • $v'(x) = 1$

   So, $f'(x) = \frac{(6)(x + 7) - (6x)(1)}{(x + 7)^2}$ Simplifying: $f'(x) = \frac{6x + 42 - 6x}{(x + 7)^2} = \frac{42}{(x + 7)^2}$

3. Evaluate the derivative at $x = 1$ to find the slope of the tangent line at the point $(1, 0.75)$:

   $f'(1) = \frac{42}{(1 + 7)^2} = \frac{42}{64} = \frac{21}{32}$

   So, the slope of the tangent line at $x = 1$ is $\frac{21}{32}$.

4. Find the equation of the tangent line: The equation of the tangent line can be written in point-slope form: $y - y_1 = m(x - x_1)$ where $m$ is the slope and $(x_1, y_1)$ is the point of tangency. Here, $m = \frac{21}{32}$, and $(x_1, y_1) = (1, 0.75)$.

   Substituting the values: $y - 0.75 = \frac{21}{32}(x - 1)$

   Simplifying: $y = \frac{21}{32}(x - 1) + 0.75$ $y = \frac{21}{32}x - \frac{21}{32} + 0.75$ Since $0.75 = \frac{24}{32}$, we get: $y = \frac{21}{32}x + \frac{3}{32}$

Thus, the equation of the tangent line is: $y = \frac{21}{32}x + \frac{3}{32}$

Would you like further clarification or more details?

Here are 5 related questions to expand the topic:

1. How do we apply the quotient rule for more complex functions?
2. How do we find the tangent line when dealing with implicit differentiation?
3. What is the geometric interpretation of the derivative in this context?
4. How would the tangent line change if we found it at a different point?
5. Can we apply linear approximation near the point $x = 1$?

Tip: The point-slope form of a line equation is a quick way to find the tangent line, especially when the slope and point are already known.