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To find the equation of the tangent line to the function \( f(x) \) at the point \( (1, f(1)) \), and to find the derivative of \( f \) at \( x = 1 \), follow these steps:

### Step 1: Determine the Slope of the Tangent Line (Derivative at \( x = 1 \))
From the graph, it appears that the tangent line at \( x = 1 \) is given as \( y = 3x - 5 \). Since the slope of a tangent line is the derivative \( f'(x) \) at that point, we conclude:
\[
f'(1) = 3
\]

### Step 2: Determine \( f(1) \)
To write the full tangent line equation, we need the point \( (1, f(1)) \). Since the tangent line is \( y = 3x - 5 \), we can substitute \( x = 1 \) into this line to find \( f(1) \):
\[
f(1) = 3(1) - 5 = 3 - 5 = -2
\]
Thus, the point of tangency is \( (1, -2) \).

### Step 3: Write the Equation of the Tangent Line
The tangent line at \( x = 1 \) is already given as \( y = 3x - 5 \). So, the equation of the tangent line is:
\[
y = 3x - 5
\]

### Summary of Answers
1. **Equation of the tangent line:** \( y = 3x - 5 \)
2. **Derivative at \( x = 1 \):** \( f'(1) = 3 \)

Would you like a detailed breakdown of the steps, or do you have any other questions?

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### Follow-Up Questions
1. How is the derivative of a function defined geometrically?
2. What are other methods to find the slope of a tangent line if it's not given directly?
3. How do we confirm the differentiability of a function at a given point?
4. What would happen if \( f(x) \) was not continuous at \( x = 1 \)?
5. Can the tangent line to a curve be horizontal? If so, what does this imply about \( f'(x) \)?

### Tip
To find the equation of a tangent line at any point \( x = a \) on \( f(x) \), calculate \( f(a) \) for the point and \( f'(a) \) for the slope, then use \( y = f'(a)(x - a) + f(a) \).

Find the equation of the line tangent to the graph of f(x) at the point (1, f(1)), then find the derivative at x = 1.

Learn how to find the equation of the tangent line to f(x) at x=1 and calculate the derivative at this point. This solution involves using the point-slope form and understanding derivatives as slopes of tangent lines, suitable for Grades 11-12.

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