Problem 17 - Calculus Early Transcendentals

Question

(a) Show that if \(f\) is differentiable on \((-\infty,+\infty)\), and if \(y=f(x)\) and \(y=f^{\prime}(x)\) are graphed in the same \(c 0\) -ordinate system, then between any two \(x\) -intercepts of \(f\) there is at least one \(x\) -intercept of \(f^{\prime}\). (b) Give some examples that illustrate this.

Step-by-Step Solution

Answer

Between any two \(x\)-intercepts of \(f\), there is an \(x\)-intercept of \(f'\). Examples: \(f(x)=\sin(x)\) and \(f(x)=x^2-1\).

Step 1: Set the context and definition

To solve this problem, consider that we have a differentiable function \(f(x)\), where the function is defined and differentiable for all real numbers \(x\). We are asked to analyze the graphs of \(y=f(x)\) and \(y=f'(x)\). In the context of this problem, an \(x\)-intercept of \(f(x)\) means \(f(x)=0\). Similarly, an \(x\)-intercept of \(f'(x)\) means the derivative \(f'(x)=0\).

Step 2: Understand Mean Value Theorem

The Mean Value Theorem (MVT) is critical here. It states that for a function \(f\) continuous on \([a, b]\) and differentiable on \((a, b)\), there exists some \(c\) in \((a, b)\) such that \(f'(c)=\frac{f(b)-f(a)}{b-a}\). This relates the value of the derivative to the slope of the secant line connecting points \((a, f(a))\) and \((b, f(b))\).

Step 3: Apply result between two \(x\)-intercepts

Let's say there are two \(x\)-intercepts of \(f(x)\) at \(x=a\) and \(x=b\) such that \(f(a)=0\) and \(f(b)=0\). By the Mean Value Theorem, there exists a point \(c\) in the interval \((a, b)\) where \(f'(c)=0\). This point \(c\) represents an \(x\)-intercept of \(f'(x)\), showing that there is at least one \(x\)-intercept of \(f'(x)\) between any two \(x\)-intercepts of \(f(x)\).

Step 4: Provide Examples

Examples help to illustrate the concept. Consider the function \(f(x) = \sin(x)\). Between any two successive \(x\)-intercepts at \(x = n\pi\) and \(x = (n+1)\pi\), the derivative \(f'(x) = \cos(x)\) has an \(x\)-intercept at \(x = \left(n + \frac{1}{2}\right)\pi\). Another example is \(f(x) = x^2 - 1\) with \(x\)-intercepts at \(x = -1\) and \(x = 1\), and the derivative \(f'(x) = 2x\) has an \(x\)-intercept at \(x = 0\).

Key Concepts

Differentiable Function

A differentiable function is a function that has a derivative at each point in its domain. In simpler terms, if you can take the derivative of a function at every possible value of its variable, it is said to be differentiable across that interval or real numbers.
The concept of differentiability is important because it ensures the function is smooth and continuous. There are no sharp corners or breaks, the function progresses without interruption.
In mathematical terms, for a function \(f(x)\) to be differentiable at a point \(x = a\), the limit:\(\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\)must exist.

• This limit gives the precise slope of the tangent to the curve at that point.
• If the function has a derivative at every point of its domain, it is said to be differentiable everywhere.

The Mean Value Theorem relies on the condition that a function must be both continuous and differentiable for it to hold true, as we examine later on how it relates to finding \(x\)-intercepts of a derivative.

x-intercept

An \(x\)-intercept of a function is where the graph of a function crosses the \(x\)-axis. At this point, the value of the function is zero. For a function \(f(x)\), the \(x\)-intercept occurs where \(f(x) = 0\).
Understanding \(x\)-intercepts is vital because they often represent significant features of the graph of a function, such as where its output changes sign.

• For polynomials, \(x\)-intercepts correspond to the roots of the equation \(f(x) = 0\).
• In the exercise context, knowing the \(x\)-intercepts of a function and its derivative is crucial. Between the \(x\)-intercepts of the function \(f(x)\), using the Mean Value Theorem, there's at least one \(x\)-intercept of \(f'(x)\).

Let's take the function \(f(x) = x^2 - 1\) as an example. It crosses the \(x\)-axis at \(x = -1\) and \(x = 1\), thus these are the function's \(x\)-intercepts. The derivative, \(f'(x) = 2x\), has its \(x\)-intercept exactly at \(x = 0\), demonstrating the theorem.

Derivative

The derivative of a function is a measure of how the function changes as its input changes. Formally, the derivative \(f'(x)\) represents the slope of the tangent line to the curve of \(f(x)\) at any given point.
Mathematically, it is expressed as the limit:\(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)This formula is the core definition of the derivative, representing instantaneous rate of change.

• The derivative helps in understanding limits, rates, and approximating function values.
• In the Mean Value Theorem context, we use the derivative to find the slope of a function between two points.

Consider the function \(f(x) = \sin(x)\). The derivative \(f'(x) = \cos(x)\) provides not only the slope of the sine function at any point but also exemplifies where its slopes hit zero or have \(x\)-intercepts, like when it crosses the \(x\)-axis.