Problem 68
Question
We know how to find the extreme values of a continuous function \(f(x)\) by investigating its values at critical points and endpoints. But what if there are no critical points or endpoints? What happens then? Do such functions really exist? Give reasons for your answers.
Step-by-Step Solution
Verified Answer
Yes, such functions exist. For example, \\( f(x) = e^x \\) on \\( (-\infty, \infty) \\).
1Step 1: Understanding the Problem
We are tasked with understanding if there are continuous functions that have no critical points or endpoints. Critical points are where the derivative is zero or undefined, and endpoints are the domain boundaries.
2Step 2: Consider Continuous Functions without Endpoints
A function without endpoints might occur on an open interval, such as the function defined on \( -\infty,\infty \) or \( (a, b) \) with no inclusion of endpoints.
3Step 3: Examine Functions Without Critical Points
Determine which continuous functions have no critical points. A critical point occurs where \( f'(x) = 0 \), so consider functions whose derivatives are never zero or undefined.
4Step 4: Identify Examples of Such Functions
One example of a continuous function without critical points on an open interval is \( f(x) = e^x \), defined on \( (-\infty, \infty) \). The derivative \( f'(x) = e^x \) is never zero.
5Step 5: Conclusion
Since \( f(x) = e^x \) is a continuous function defined over an open interval and has no critical points, it is indeed possible for such functions to exist.
Key Concepts
Critical PointsContinuous FunctionsDerivative
Critical Points
Critical points are special values of the variable in a function where the derivative is either zero or does not exist. These points often indicate where a function's slope levels off, peaks, or dips. They can help us find local maximums, minimums, or points of inflection in a function.
Finding critical points involves taking the derivative of a function and setting it equal to zero. If the derivative does not exist at certain points, these are also considered critical points.
Finding critical points involves taking the derivative of a function and setting it equal to zero. If the derivative does not exist at certain points, these are also considered critical points.
- Critical points help determine where a function changes direction.
- Understanding critical points is essential for finding the extremum of functions.
- They provide insight into the behavior of a function over its domain.
Continuous Functions
Continuous functions are functions where small changes in the input result in small changes in the output. This continuity means there are no "jumps" or breaks as you proceed along the function. A continuous function is defined for every single point in its domain.
For a function \( f(x) \) to be continuous, it must meet these conditions:
For a function \( f(x) \) to be continuous, it must meet these conditions:
- The function must be defined at every point in the interval.
- The function's limit as it approaches a point from both directions must be equal to the value of the function at that point.
- There should be no abrupt changes in the function's output.
Derivative
The derivative of a function measures how the function's output changes concerning changes in the input. It is often described as the "rate of change" or "slope" of the function. Understanding derivatives is crucial in calculus for defining the behavior and direction of a function.
The derivative \( f'(x) \) of a function \( f(x) \) represents:
The derivative \( f'(x) \) of a function \( f(x) \) represents:
- The instantaneous rate of change at any given point.
- The slope of the tangent line to the curve at a point.
- How rapidly or slowly a function is increasing or decreasing.
Other exercises in this chapter
Problem 67
Odd functions If an odd function \(g(x)\) has a local minimum value at \(x=c,\) can anything be said about the value of \(g\) at \(x=-c ?\) Give reasons for you
View solution Problem 68
Solve the initial value problems in Exercises \(67-86\). $$ \frac{d y}{d x}=10-x, \quad y(0)=-1 $$
View solution Problem 68
Sketch a smooth connected curve \(y=f(x)\) with $$ \begin{array}{ll}{f(-2)=8,} & {f^{\prime}(2)=f^{\prime}(-2)=0} \\ {f(0)=4,} & {f^{\prime}(x) 0 \text { for }|
View solution Problem 69
Solve the initial value problems in Exercises \(67-86\). $$ \frac{d y}{d x}=\frac{1}{x^{2}}+x, \quad x>0 ; \quad y(2)=1 $$
View solution