Question

If \(f:[a,b] \to [c,d]\) is a continuous and strictly increasing function, then \(\frac{d-c}{b-a}\) is

• A. Value of the function at a point \(t \in (a,b)\)
• B. Value of the function at \(t \in (a,b)\) such that \(f'(t)=0\)
• C. Slope of the tangent drawn to the curve \(y=f(t)\) at a point \(t \in (c,d)\)
• D. Slope of the tangent drawn to the curve \(y=f(t)\) at a point \(t \in (a,b)\)

Hint:

The expression \(\frac{f(b)-f(a)}{b-a}\) represents the average rate of change over the interval. The Mean Value Theorem guarantees there is an instantaneous rate of change (slope of tangent) equal to this average.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The question asks for the geometric interpretation of the expression \(\frac{d-c}{b-a}\) given the properties of the function \(f\).
Step 2: Applying Function Properties:
Since \(f\) is a strictly increasing function from \([a,b]\) to \([c,d]\), it maps the smallest input to the smallest output and the largest input to the largest output. Therefore: \(f(a) = c\) \(f(b) = d\)
Step 3: Interpreting the Expression:
Substitute these values into the expression: \[ \frac{d-c}{b-a} = \frac{f(b) - f(a)}{b - a} \] This ratio represents the slope of the secant line joining the points \((a, f(a))\) and \((b, f(b))\).
Step 4: Using the Mean Value Theorem (Lagrange's MVT):
Since \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\) (implied by the context of "tangent" in options), by the Mean Value Theorem, there exists at least one point \(t \in (a,b)\) such that: \[ f'(t) = \frac{f(b) - f(a)}{b - a} \] Here, \(f'(t)\) represents the slope of the tangent to the curve \(y=f(x)\) at the point \(x=t\). Thus, \(\frac{d-c}{b-a}\) is the slope of the tangent drawn to the curve \(y=f(t)\) at a point \(t \in (a,b)\).