Problem 68 - Thomas Calculus

Question

Suppose that \(f\) has a negative derivative for all values of \(x\) and that \(f(1)=0 .\) Which of the following statements must be true of the function $$ h(x)=\int_{0}^{x} f(t) d t ? $$ Give reasons for your answers. a. \(h\) is a twice-differentiable function of \(x\) b. \(h\) and \(d h / d x\) are both continuous. c. The graph of \(h\) has a horizontal tangent at \(x=1\) . d. \(h\) has a local maximum at \(x=1\) e. \(h\) has a local minimum at \(x=1\) f. The graph of \(h\) has an inflection point at \(x=1\) . g. The graph of \(d h / d x\) crosses the \(x\) -axis at \(x=1\) .

Step-by-Step Solution

Answer

h is twice-differentiable and continuous. h has a local minimum at x=1, with a horizontal tangent, but no inflection point.

Step 1: Understanding the Function h(x)

The function given is \( h(x) = \int_{0}^{x} f(t) \, dt \). According to the Fundamental Theorem of Calculus, the derivative of \( h \) with respect to \( x \) is \( \frac{dh}{dx} = f(x) \). This indicates that \( h \) is differentiable wherever \( f(x) \) is defined.

Step 2: Analyzing the Continuity of h and dh/dx

Since \( f(x) \) is continuous and \( h(x) \) is derived from an integral with a continuous integrand, it follows that \( h \) and its derivative \( \frac{dh}{dx} = f(x) \) are both continuous.

Step 3: Evaluating Twice Differentiability

For twice differentiability, we need \( f(x) \) to be differentiable so that \( \frac{d^2h}{dx^2} = f'(x) \) exists. The problem states that \( f \) has a negative derivative, thus \( f \) is differentiable, implying that \( h(x) \) is twice differentiable.

Step 4: Checking if the Graph of h has a Horizontal Tangent at x=1

A horizontal tangent occurs where \( \frac{dh}{dx} = 0 \). At \( x=1 \), \( \frac{dh}{dx} = f(1) \). Given that \( f(1)=0 \), \( \frac{dh}{dx} = 0 \) at \( x=1 \). Thus, the graph of \( h \) has a horizontal tangent at \( x=1 \).

Step 5: Determining Local Maxima and Minima at x=1

At \( x=1 \), we have \( \frac{dh}{dx} = 0 \) but \( f(x) \) has a negative derivative. For a local maximum, \( f \) must increase (\( f'(x) > 0 \)), but it decreases here (\( f'(x) < 0 \)), ruling out a local max. Consequently, \( h(x) \) has a local minimum at \( x=1 \).

Step 6: Checking for an Inflection Point at x=1

An inflection point occurs where the second derivative changes sign. Since \( f(x) \) has a negative derivative everywhere, \( f'(x) < 0 \) indicates no sign change at \( x=1 \). Therefore, there is no inflection point at \( x=1 \).

Step 7: Analyzing where dh/dx crosses the x-axis

The graph of \( \frac{dh}{dx} = f(x) \) crossing the \( x \)-axis at \( x=1 \) means \( f(1)=0 \). We know \( f(1)=0 \), confirming that the graph of \( \frac{dh}{dx} \) does indeed cross the \( x \)-axis at \( x=1 \).

Key Concepts

Continuity of Functions

Continuity in functions means that you can draw the graph of the function without lifting your pencil.
It is essential for functions that are based on integrals, like the function \( h(x) = \int_{0}^{x} f(t) \, dt \) we are working with here.
Since \( f(x) \) is continuous, any integral involving \( f(x) \) will also be continuous.

• This is due to the fact that the Fundamental Theorem of Calculus assures us that the integral of a continuous function is continuous.
• This carries over to both \( h(x) \) and its derivative \( \frac{dh}{dx} = f(x) \), ensuring both are smooth without any jumps or breaks.

This property guarantees smooth transitions in the function's behavior, crucial for analyzing and predicting other properties like differentiability and local extrema.

Local Extrema

Local extrema are simply the highest or lowest points in a certain section of a function's graph.
In mathematical terms, these are referred to as local maxima or minima.
For the function \( h(x) \), determining local extrema involves the derivative \( \frac{dh}{dx} = f(x) \).

• A local maximum occurs at a point if the derivative changes from positive to negative there.
• A local minimum occurs if the derivative changes from negative to positive.

Given that \( f(x) \) has a negative derivative, indicating it's always decreasing, \( h \) doesn't have places where \( f \) goes from positive to negative or vice versa.
However, at \( x = 1 \), \( \frac{dh}{dx} = 0 \) and \( f \) is decreasing, leading to a local minimum rather than a maximum.
Understanding local extrema requires focusing heavily on the behavior of the derivative across the function's domain.

Differentiability

Differentiability of a function refers to its capability to have a derivative everywhere within a given interval.
This feature is essential for smooth and predictable function behavior, particularly if we're interested in analyzing rates of change or finding optimal points.

• For \( h(x) = \int_{0}^{x} f(t) \, dt \), differentiability is guaranteed wherever \( f(x) \) is continuous.
• It's confirmed here because we know that \( f(x) \) is not only continuous but also differentiable, with respect to the given conditions in the problem.

In this case, \( h(x) \) is not just once, but twice differentiable because \( f(x) \) itself has a defined negative derivative everywhere.
This means we can find both \( \frac{dh}{dx} \) and further \( \frac{d^2h}{dx^2} \).
Differentiability ascertains the existence of tangents, thus influencing how we determine the nature of \( h(x) \) at specific points, such as its behavior at \( x = 1 \).