Problem 78
Question
Prove that if \(f^{\prime}(x)=0\) for all \(x\) in an interval \((a, b),\) then \(f\) is constant on \((a, b) .\)
Step-by-Step Solution
Verified Answer
If the derivative of a function is zero for all points in an interval, the function is constant over that interval. This is derived from the Mean Value Theorem. Therefore, if \( f'(x) = 0 \) for all \( x \) in an interval (a, b), it follows that the function \( f \) is constant over this interval.
1Step 1: State The Mean Value Theorem
The Mean Value Theorem states: If a function \( f \) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number \( c \) in the interval (a, b) such that the derivative of the function at that point \( c \) equals the average rate of change over the interval [a, b]. In mathematical terms, there exists some \( c \) in (a, b) such that \( f'(c) = \frac{f(b)-f(a)}{b-a} \).
2Step 2: Supposition and Implication of \( f'(x) = 0 \)
Since the derivative \( f'(x) \) of the function \( f \) equals zero for all \( x \) in the interval (a, b), we can suppose \( f'(c) = 0 \) for any \( c \) in the interval (a, b). Using the Mean Value Theorem, this implies that \( \frac{f(b)-f(a)}{b-a} = 0\). Simplification leads to \( f(b) = f(a) \).
3Step 3: Proof of the Function Being Constant
From the previous step, for any two points \( a,b \) in the interval (a, b), we have \( f(a) = f(b) \). This is the definition of a constant function on the interval (a, b). Hence, it can be concluded that the function \( f \) is constant on the interval (a, b).
Key Concepts
DifferentiabilityConstant FunctionZero Derivative
Differentiability
In mathematics, differentiability is a crucial concept when dealing with functions. When we say a function is differentiable, we mean that it has a derivative at every point in its domain. Specifically, for a function to be differentiable on an interval, it must be smooth and have no sharp turns or breaks. It ensures that the function behaves in a predictable way everywhere in that interval.
For example, if you look at a graph of a differentiable function, you should be able to draw a tangent line at each point without lifting your pencil. This property of differentiability is essential because it allows us to apply powerful mathematical tools, such as the Mean Value Theorem. It also guarantees that Newton's methods and other advanced calculus techniques work effectively in finding solutions.
For example, if you look at a graph of a differentiable function, you should be able to draw a tangent line at each point without lifting your pencil. This property of differentiability is essential because it allows us to apply powerful mathematical tools, such as the Mean Value Theorem. It also guarantees that Newton's methods and other advanced calculus techniques work effectively in finding solutions.
- If a function is differentiable, it must also be continuous (though the reverse isn't always true).
- Differentiability implies the existence of a smooth slope at every point.
Constant Function
A constant function is one of the simplest types of functions you can encounter. When a function is constant, it means its value does not change no matter what input you use from its domain. To put it simply, as you move along the x-axis, the output (or y-value) remains the same.
Mathematically, if you graph a constant function, it will be a straight horizontal line. This means there is no change or slope, making the derivative always 0. This simplicity is powerful in contexts like the Mean Value Theorem, where recognizing a function as constant can help determine critical properties. If a function's derivative is zero over an interval, the function does not "move" or "change," staying constant.
Mathematically, if you graph a constant function, it will be a straight horizontal line. This means there is no change or slope, making the derivative always 0. This simplicity is powerful in contexts like the Mean Value Theorem, where recognizing a function as constant can help determine critical properties. If a function's derivative is zero over an interval, the function does not "move" or "change," staying constant.
- In a constant function, \[ f(x) = c \] where \( c \) is a constant value.
- Generally shown as a horizontal line on a graph.
Zero Derivative
The concept of a zero derivative is fundamental in calculus. When the derivative of a function \( f'(x) \) is zero, it indicates that the function's slope or rate of change is zero at that point. Simply put, there is no increase or decrease; the function "rests."
A critical implication of a zero derivative over an interval is that it leads to the identification of constant functions. According to the Mean Value Theorem, if \( f'(x) = 0 \) for every \( x \) in \((a, b)\), it means there are no differences in the function values across the interval, hence \( f(a) = f(b) \).
A critical implication of a zero derivative over an interval is that it leads to the identification of constant functions. According to the Mean Value Theorem, if \( f'(x) = 0 \) for every \( x \) in \((a, b)\), it means there are no differences in the function values across the interval, hence \( f(a) = f(b) \).
- Zero derivative signifies zero slope — the function is flat.
- Helps in concluding that a function must be constant over the interval.
Other exercises in this chapter
Problem 78
Analyzing a Graph Using Technology In Exercises \(75-82,\) use a computer algebra system to analyze the graph of the function. Label any extrema and/or asymptot
View solution Problem 78
Numerical, Graphical, and Analytic Analysis Consider the functions \(f(x)=x\) and \(g(x)=\sin x\) on the interval \((0, \pi) .\) (a) Complete the table and make
View solution Problem 79
Analyzing a Graph Using Technology In Exercises \(75-82,\) use a computer algebra system to analyze the graph of the function. Label any extrema and/or asymptot
View solution Problem 79
Proof In Exercises 79 and \(80,\) let \(f\) and \(g\) represent differentiable functions such that \(f^{\prime \prime} \neq 0\) and \(g^{\prime \prime} \neq 0\)
View solution