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 $x = 1$: $f'(1) = 3$

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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 $y = f'(a)(x - a) + f(a)$.