Problem 67 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Statements a and b are true; c, d, e, f, and g are false.

Step 1: Analyzing differentiability and continuity

Function \(g(x)\) is defined as \(\int_{0}^{x} f(t)\, dt \). Since \(f(x)\) has a positive derivative everywhere, it means \(f(x)\) is continuous. Therefore, \(g(x)\), an integral of a continuous function, is both continuous and differentiable. Statements **a** and **b** are true.

Step 2: Calculating the derivative

The derivative of \(g(x)\) with respect to \(x\) is obtained from the Fundamental Theorem of Calculus: \(\frac{dg}{dx} = f(x)\). For \(x = 1\), since \(f(x)\) is positive, \(\frac{dg}{dx}\) is non-zero, so the graph of \(g(x)\) does not have a horizontal tangent at \(x = 1\). Statement **c** is false.

Step 3: Determining local extrema

For \(g\) to have a local extremum at \(x = 1\), \(\frac{dg}{dx} = 0\). Since \(\frac{dg}{dx} = f(1) > 0\), no extremum exists at \(x = 1\). Therefore, both statements **d** and **e** are false.

Step 4: Checking for an inflection point

For an inflection point, \(\frac{d^2g}{dx^2}\) must change sign. \(\frac{d^2g}{dx^2} = f'(x)\). Given \(f'(x) > 0\) for all \(x\), \(g(x)\) cannot have an inflection point at \(x = 1\). Statement **f** is false.

Step 5: Analyzing the behavior of dg/dx

Since \(\frac{dg}{dx} = f(x)\) is never zero (as \(f(x) > 0\)), the function \(\frac{dg}{dx}\) does not cross the \(x\)-axis at \(x = 1\). Statement **g** is false.

Key Concepts

Differentiability

Differentiability is a crucial concept in calculus which helps us understand if a function has a derivative at every point in its domain. If a function is differentiable at a point, it means the function is smooth and has no sharp corners or breaks at that point. Differentiability implies continuity, but not vice versa; a function can be continuous but not differentiable.

In the context of the exercise, the function \(g(x) = \int_{0}^{x} f(t) \, dt\) is differentiable because \(f\) is a positive derivative for all \(x\) values, indicating no interruptions or abrupt changes. Due to this property, the derivative of \(g(x)\) exists and can be expressed through the Fundamental Theorem of Calculus as \(\frac{dg}{dx} = f(x)\). This means \(g(x)\) is smooth and differentiable across its domain.

Continuity

Continuity is the property of a function where small changes in the input result in small changes in the output. A continuous function has no gaps, jumps, or interruptions. If a function is differentiable across its domain, it is also continuous throughout that domain.

In the given problem, since \(f(x)\) has a positive derivative everywhere, we conclude that \(f(x)\) itself must be continuous. As \(g(x)\) is an integral of a continuous function \(f(x)\), it inherits this continuity property. Thus, \(g(x)\) is continuous; you won't find any gaps or abrupt jumps when analyzing \(g(x)\). This makes it easier to evaluate and study \(g\) at any point within its range.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is central to understanding the relationship between differentiation and integration—two core operations in calculus. It consists of two main parts:

• First Part: It connects the integral of a function with its antiderivative. Specifically, if \(F(x)\) is an antiderivative of \(f(x)\), then the integral of \(f(t)\) from \(a\) to \(b\) is \(F(b) - F(a)\).
• Second Part: It states that if \(f\) is continuous over an interval and \(G(x) = \int_{a}^{x} f(t) \, dt\), then \(G\) is continuous and differentiable on that interval, with its derivative being \(f(x)\).

These properties allow us to solve this exercise more easily. By applying the theorem, we determine the derivative of \(g(x)\) with respect to \(x\) is \(f(x)\). Since \(f(x)\) is always positive, \(g(x)\) remains differentiable, and \(\frac{dg}{dx}\) is non-zero across the domain. This means no horizontal tangents, local maxima, or minima where you might expect non-continuous derivatives or changes in concavity to occur strictly at \(x=1\).