Problem 46
Question
Assume that \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\). Show
that if \(f(a)
Step-by-Step Solution
Verified Answer
By the Mean Value Theorem, if \( f(a) < f(b) \), there is a \( c \in (a, b) \) such that \( f'(c) > 0 \).
1Step 1: Understand the Mean Value Theorem (MVT)
The Mean Value Theorem states that if a function \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one point \( c \) in \( (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). This sets the groundwork for finding a point where the derivative is positive.
2Step 2: Apply the Conditions of MVT
We know that \( f(a) < f(b) \). Thus, \( f(b) - f(a) > 0 \). Since \( b - a > 0 \) (as \( a eq b \)), the expression \( \frac{f(b) - f(a)}{b - a} \) is positive. Therefore, by MVT, there is at least one \( c \) in \( (a, b) \) where \( f'(c) > 0 \).
3Step 3: Conclusion from MVT Application
By applying the Mean Value Theorem to the given conditions and the fact that \( f(a) < f(b) \), we conclude that \( f' \) is positive at some point between \( a \) and \( b \). This means there is a point \( c \) where the slope of the tangent (the derivative) is positive.
Key Concepts
ContinuityDifferentiabilityDerivativeInterval
Continuity
In mathematics, **continuity** is a property of a function where small changes in the input result in small changes in the output. Imagine a graph that you can draw without lifting your pencil off the paper; that's a continuous function. For the Mean Value Theorem to apply, the function must be continuous over the closed interval \( [a, b]\). This ensures there are no breaks or jumps in the graph on this interval.
Here's why that's crucial:
Here's why that's crucial:
- It allows the function to smoothly transition between points \(a\) and \(b\).
- Ensures that every value between \(f(a)\) and \(f(b)\) is hit by the function.
Differentiability
**Differentiability** takes continuity a step further by considering if the derivative exists at every point in an interval. For a function to be differentiable on an open interval \( (a, b)\), its graph must be smooth without any sharp corners or cusps within this range. Why does differentiability matter?
Because:
Because:
- It guarantees that the derivative, or the slope of the tangent, can be calculated accurately at any point \(c\) within \((a, b)\).
- It ensures the transition between points is smooth enough to apply the Mean Value Theorem.
Derivative
The **derivative** of a function at a given point tells us the slope of the tangent to the function's graph at that point. In practical terms, it's like asking, "How steep is my hill at \(c\)?" For the Mean Value Theorem, we're interested in finding such a point \(c\) where the derivative meets certain conditions.
Specifically:
Specifically:
- If \(f(a) < f(b)\), then somewhere between \(a\) and \(b\), the derivative \(f'(c)\) must be positive.
- This means the slope at \(c\) is uphill, indicating a positive change over the interval.
Interval
An **interval** refers to the range of inputs under consideration, denoted here as \([a, b]\) and \((a, b)\). Knowing the type of interval helps in determining the applicable rules for analysis.
- \([a, b]\) is a closed interval, including both endpoints \(a\) and \(b\).
- \((a, b)\) is open, excluding these endpoints, tending to focus on the interior points.
Other exercises in this chapter
Problem 46
Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow \infty}\left(1+\frac{5}{x}\right)^{x} $$
View solution Problem 46
In Problems \(41-46\), assume that \(a\) is a positive constant. Find the general antiderivative of the given function. $$ f(x)=\frac{e^{-a x}+e^{a x}}{2 a} $$
View solution Problem 47
In Problems 47-58, find the general solution of the differential equation. $$ \frac{d y}{d x}=\frac{2}{x}-x, x>0 $$
View solution Problem 47
Assume that \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\). Assume further that \(f(a)=f(b)=0\) but \(f\) is not constant on \([a, b] .\) Ex
View solution