The graph of function f contains the point `P(1, 2) and Q(s, r)`. The equation of the secant line through P and Q is `y=((s^2+2s-3)/(s-1))x-1-s`. The value of `f'(1)`, is

Options

A. (a)2
B. (b)3
C. (c)4
D. (d)non-existent

Answer

To find the value of \( f'(1) \) given the secant line through points \( P(1, 2) \) and \( Q(s, r) \), we will follow these steps:

### Step 1: Identify the slope of the secant line
The equation of the secant line is given by:

\[ y = \frac{s^2 + 2s - 3}{s - 1} x - 1 - s \]

From this equation, we can identify the slope \( m \) of the secant line as:

\[ m = \frac{s^2 + 2s - 3}{s - 1} \]

### Step 2: Calculate the derivative at \( x = 1 \)
To find \( f'(1) \), we need to evaluate the limit of the slope as \( s \) approaches 1:

\[ f'(1) = \lim_{s \to 1} \frac{s^2 + 2s - 3}{s - 1} \]

### Step 3: Substitute \( s = 1 \)
Substituting \( s = 1 \) directly into the expression gives us:

\[ f'(1) = \frac{1^2 + 2(1) - 3}{1 - 1} = \frac{0}{0} \]

This is an indeterminate form, so we will apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
Using L'Hôpital's Rule, we differentiate the numerator and the denominator:

- The derivative of the numerator \( s^2 + 2s - 3 \) is \( 2s + 2 \).
- The derivative of the denominator \( s - 1 \) is \( 1 \).

Thus, we have:

\[ f'(1) = \lim_{s \to 1} \frac{2s + 2}{1} \]

### Step 5: Evaluate the limit
Now substituting \( s = 1 \):

\[ f'(1) = 2(1) + 2 = 4 \]

### Conclusion
Therefore, the value of \( f'(1) \) is:

\[ \boxed{4} \]