Let ff be a differentiable function with f, of, 1, equals, 1f(1)=1 and f, prime, of, 1, equals, 2, .f 
′
 (1)=2. Let the function g, of, x, equals, left parenthesis, 3, x, plus, 4, right parenthesis, f, of, x, .g(x)=(3x+4)f(x). Write the equation of the line tangent to the graph of gg at the point where x, equals, 1, .x=1.

We are given that \( g(x) = (3x + 4)f(x) \), and we need to find the equation of the tangent line to the graph of \( g(x) \) at \( x = 1 \). To do so, we need two things:

1. The value of \( g(1) \), which will give us the point of tangency.
2. The value of \( g'(1) \), which will give us the slope of the tangent line.

### Step 1: Compute \( g(1) \)

We know that \( g(x) = (3x + 4)f(x) \) and we are given that \( f(1) = 1 \). Therefore, at \( x = 1 \):

\[
g(1) = (3(1) + 4)f(1) = (3 + 4)(1) = 7.
\]

Thus, the point of tangency is \( (1, 7) \).

### Step 2: Compute \( g'(x) \)

Using the product rule, we differentiate \( g(x) = (3x + 4)f(x) \):

\[
g'(x) = (3x + 4)f'(x) + 3f(x).
\]

Now, substituting \( x = 1 \), \( f(1) = 1 \), and \( f'(1) = 2 \):

\[
g'(1) = (3(1) + 4)f'(1) + 3f(1) = (3 + 4)(2) + 3(1) = 7(2) + 3 = 14 + 3 = 17.
\]

Thus, the slope of the tangent line at \( x = 1 \) is 17.

### Step 3: Equation of the tangent line

The equation of the tangent line is given by the point-slope form of a line:

\[
y - y_1 = m(x - x_1),
\]

where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency. Substituting \( m = 17 \), \( x_1 = 1 \), and \( y_1 = 7 \):

\[
y - 7 = 17(x - 1).
\]

Simplifying:

\[
y = 17(x - 1) + 7 = 17x - 17 + 7 = 17x - 10.
\]

Thus, the equation of the tangent line is:

\[
y = 17x - 10.
\]

Let me know if you want more details or have any questions!

### Related Questions:
1. How does the product rule for differentiation apply in this context?
2. What if \( f(1) \) were different, how would the tangent line change?
3. How do you interpret the slope of 17 geometrically?
4. How does knowing the derivative \( f'(x) \) at a point influence the graph's behavior?
5. Can the same method be used for any product of two functions?

### Tip:
When using the product rule, make sure to carefully differentiate both the first and second functions, and remember to apply the chain rule if necessary!

Let f be a differentiable function with f(1) = 1 and f'(1) = 2. Let the function g(x) = (3x + 4)f(x). Write the equation of the line tangent to the graph of g at the point where x = 1.

Find the equation of the tangent line to the function g(x) = (3x + 4)f(x) at x = 1 using the product rule and point-slope form. Learn how to differentiate composite functions and apply calculus principles to solve this problem. Suitable for Grades 11-12 calculus students.

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