Problem 69
Question
Assume that \(f\) is differentiable on \(a \leq x \leq b\) and that \(f(b)
Step-by-Step Solution
Verified Answer
There exists a point where the derivative is negative by the Mean Value Theorem.
1Step 1: Understand the Problem Statement
We are given a differentiable function \(f\) on the interval \([a, b]\) such that \(f(b) < f(a)\). Our goal is to demonstrate that there exists at least one point \(c\) in the interval \((a, b)\) where the derivative \(f'(c) < 0\).
2Step 2: Apply the Mean Value Theorem
According to the Mean Value Theorem for derivatives, if a function \(f\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \(c\) in \((a, b)\) such that:\[f'(c) = \frac{f(b) - f(a)}{b - a}\]
3Step 3: Analyze the Expression \(f(b) - f(a)\)
We are given that \(f(b) < f(a)\), meaning that \(f(b) - f(a) < 0\). This implies that the quotient \(\frac{f(b) - f(a)}{b - a}\) is negative since \(b - a > 0\) for \(a < b\).
4Step 4: Conclude Using the Mean Value Theorem
By the Mean Value Theorem, there exists some point \(c\) in \((a, b)\) such that the derivative \(f'(c)\) is equal to the negative expression \(\frac{f(b) - f(a)}{b - a}\). Thus, \(f'(c) < 0\).
Key Concepts
Differentiable FunctionDerivativesContinuous Function
Differentiable Function
A differentiable function is a function that has a derivative at every point within its domain. In simple terms, it means you can find the slope of the tangent to the function's graph at any point where the function is defined. This is an important concept because it ensures that the function behaves "smoothly" without any abrupt jumps or sharp turns.
To determine if a function is differentiable, you need to check two main conditions:
To determine if a function is differentiable, you need to check two main conditions:
- The function must be continuous at the point in question. This means there should be no breaks or holes in the graph of the function at that point.
- The derivative must exist at that point. This means you should be able to calculate the slope of the tangent line using the limit process.
Derivatives
Derivatives are a fundamental concept in calculus representing the rate of change of a function. A derivative gives the slope of the function at any given point, telling us how the function is increasing or decreasing at that precise location. Using derivatives, you can identify various properties of the function, like local maxima, minima, and points where the function's behavior changes.
Here's how a derivative is generally defined:
Here's how a derivative is generally defined:
- The derivative of a function \(f\) at a point \(x\) is given by the limit:\[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}\]
- This expression calculates the slope of the tangent line to the function at the point \(x\).
- The derivative tells us how the function \(f\) value changes as the input \(x\) is slightly altered.
Continuous Function
A continuous function is one that can be drawn without lifting your pencil from the paper. For a function to be continuous at a point, the function's value must match its limit approaching that point. Continuity is a necessary precondition for a function to be differentiable. If a function is not continuous at a point, it cannot be differentiable there.
There are three types of continuity:
There are three types of continuity:
- Pointwise Continuity: a function is continuous at a point if the left-hand and right-hand limits at that point equal the function's value.
- Local Continuity: a function is continuous in an interval if it is continuous at each point within that interval.
- Global Continuity: when continuity holds over the entire domain of the function.
Other exercises in this chapter
Problem 69
Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x^{2 / 3}(x+2)$$
View solution Problem 69
Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $
View solution Problem 69
Sketch the graph of a continuous function \(y=g(x)\) such that a. \(g(2)=2,02,\) and \(g^{\prime}(x) \rightarrow-1^{+}\) as \(x \rightarrow 2^{+}\) b. \(g(2)=2,
View solution Problem 69
Try it-you just keep on cycling. Find the limits some other way. $$\lim _{x \rightarrow(\pi / 2)} \frac{\sec x}{\tan x}$$
View solution