Problem 18 - A Textbook of Differential Calculus

Question

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 \(y=\left(\frac{s^{2}+2 s-3}{s-1}\right) x-1-s\). The value of \(f^{\prime}(1)\), is (a) 2 (b) 3 . (c) 4 (d) Non existent

Step-by-Step Solution

Answer

The value of \(f'(1)\) is 4, so the answer is (c).

Step 1: Understanding the Secant Line Equation

The given equation of the secant line is \(y = \left(\frac{s^2 + 2s - 3}{s-1}\right)x - 1 - s\). This equation passes through points \(P(1, 2)\) and \(Q(s, r)\). The slope of this secant line is \(\frac{s^2 + 2s - 3}{s-1}\).

Step 2: Relate the Secant Line Slope to Derivative

The slope of the secant line between points \((1, 2)\) and \((s, f(s))\) is \(\frac{f(s) - f(1)}{s-1}\). We are given that \(P(1, 2)\), thus \(f(1) = 2\).

Step 3: Compare Secant Line Slope with Function Slope

Equate the slope from the secant line equation to the difference quotient definition: \[ \frac{s^2 + 2s - 3}{s-1} = \frac{f(s) - 2}{s-1} \]. This equation should be true for all \(s\) close to 1.

Step 4: Simplify for a Derivative

For the derivative \(f'(1)\) to exist, we need the left-hand side \( \frac{s^2 + 2s - 3}{s-1} \) to give a finite value as \(s\) approaches 1. Let's factor the numerator: \(s^2 + 2s - 3 = (s - 1)(s + 3)\).

Step 5: Cancel Common Factors

Now, cancel the common \((s - 1)\) factor: \[ \frac{(s-1)(s+3)}{s-1} = s + 3 \] whenever \(s eq 1\).

Step 6: Evaluate the Limit for Derivative

Taking the limit as \(s\) approaches 1 gives: \(f'(1) = \lim_{{s \to 1}} (s + 3) = 1 + 3 = 4\). This is the derivative at the point.

Key Concepts

Secant Line

The concept of a secant line is fundamental in differential calculus. A secant line connects two points on a function's curve. In our exercise, these points are \(P(1, 2)\) and \(Q(s, r)\). This line allows us to measure the average rate of change of the function between these two points.

The equation of the secant line in this context is given as \( y = \left(\frac{s^2 + 2s - 3}{s-1}\right)x - 1 - s \). Here, the expression \( \frac{s^2 + 2s - 3}{s-1} \) represents the slope of the secant line.

Understanding the secant line is essential as it forms the basis for determining an instantaneous rate of change or the derivative.

Derivative

In calculus, the derivative represents the instantaneous rate of change of a function. While the secant line gives us an average rate of change, the derivative refines this by considering the limit as the interval approaches zero.

The exercise asks us to find \(f'(1)\), the derivative of the function at \(x = 1\). Derivatives are a core concept in calculus and are used to compute the slope of the tangent line at a specific point on a curve.

To find \(f'(1)\), we leverage the slope of the secant line and simplify it using limits as \(s\) approaches 1. This way, we transition from the average rate provided by the secant to the instantaneous rate, or derivative, at the specific point.

Function Slope

The slope of a function at a specific point is crucial for understanding its behavior. The slope tells us how steep the function is at a particular point and is given by its derivative at that point.

For the points \(P(1, 2)\) and \(Q(s, r)\) in our exercise, the slope of the secant line, \( \frac{s^2 + 2s - 3}{s-1} \), transitions to the derivative \(f'(1)\) when we take the limit as \(s\) approaches 1.

• The secant line slope provides the average rate of change between two points.
• The function's slope at a point (derivative) gives us the instantaneous rate of change at that specific point.

Understanding these concepts helps in determining how the function increases or decreases at any given point.

Difference Quotient

The difference quotient is a formula that gives us the slope of the secant line between two points of a function. It is defined as \( \frac{f(s) - f(a)}{s - a}\) where \(s\) and \(a\) are distinct points on the function.

In our scenario, the difference quotient is set up as \( \frac{f(s) - 2}{s-1} \) since \( f(1) = 2 \). This expression is crucial because it lays the groundwork for computing the derivative.

We equate this to the given secant line slope \( \frac{s^2 + 2s - 3}{s-1} \). Simplifying the expression by factoring enables us to cancel terms and compute the derivative as a limit: \

\[ f'(1) = \lim_{{s \to 1}} (s + 3) = 4 \]The difference quotient is foundational for transitioning from average rate of change to the precise instantaneous rate of change in calculus.