91Ó°ÊÓ

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. \(\begin{array}{l}{\text { a. } g \text { is a differentiable function of } x \text { . }} \\ {\text { b. } g \text { is a continuous function of } x \text { . }} \\ {\text { c. The graph of } g \text { has a horizontal tangent at } x=1 \text { . }} \\ {\text { d. } g \text { has a local maximum at } x=1 \text { . }}\\\\{\text { e. } g \text { has a local minimum at } x=1.} \\\ {\text { f. The graph of } g \text { has an inflection point at } x=1.} \\\ {\text { g. The graph of } d g / d x \text { crosses the } x \text { -axis at } x=1}.\end{array}\)

Step by step solution

01

Analyzing Statement a

Determine if \( g(x) = \int_{0}^{x} f(t) \, dt \) is differentiable. By the Fundamental Theorem of Calculus, if \( f \) is continuous, then \( g \) is differentiable everywhere and \( g'(x) = f(x) \). Since \( f(t) \) is given with a positive derivative, it implies \( f(x) \) is continuous. Therefore, \( g(x) \) is differentiable for all \( x \).

02

Analyzing Statement b

Since \( g(x) \) is the integral of a continuous function, it must be continuous itself. A differentiable function, like \( g(x) \), is also continuous. Therefore, \( g(x) \) is a continuous function.

03

Analyzing Statement c

The graph of \( g(x) \) has a horizontal tangent at a point if \( g'(x) = 0 \) at that point. From step 1, \( g'(x) = f(x) \). Since \( f(x) \) has a positive derivative, \( f(x) eq 0 \) for all \( x \), implying the graph of \( g \) never has a horizontal tangent.

04

Analyzing Statement d

For \( g(x) \) to have a local maximum, \( g'(x) \) should change from positive to negative at \( x=1 \). Since \( g'(x) = f(x) > 0 \) for all \( x \), \( g(x) \) cannot have a local maximum at \( x=1 \).

05

Analyzing Statement e

For \( g(x) \) to have a local minimum, \( g'(x) \) should change from negative to positive at \( x=1 \). Again, \( g'(x) = f(x) > 0 \) for all \( x \), so \( g(x) \) cannot have a local minimum at \( x=1 \).

06

Analyzing Statement f

An inflection point on \( g(x) \) would require the second derivative \( g''(x) \) to change sign at \( x=1 \). We need the given \( f' \) (which is positive) to change sign, but since it is positive everywhere, \( g(x) \) has no inflection point at \( x=1 \).

07

Analyzing Statement g

The graph of \( \frac{dg}{dx} = f(x) \) crossing the x-axis would imply \( f(x) = 0 \) at \( x=1 \). However, since \( f(x) > 0 \) everywhere, \( \frac{dg}{dx} \) doesn't cross the x-axis at \( x=1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiability of Integrals

When we discuss the differentiability of integrals, we are directly referring to the Fundamental Theorem of Calculus. This theorem provides a bridge between differentiation and integration, key concepts in calculus. It says that if a function \(f\) is continuous on a closed interval, then the function \(g(x) = \int_{0}^{x} f(t) \, dt\) is differentiable, and its derivative \(g'(x)\) is exactly \(f(x)\).
This means for our function \(g\), derived from the continuous function \(f\), it will be differentiable over its entire domain. In simpler terms, the smoothness of \(f\), ensures that \(g\) itself won't have any sharp turns or corners. Differentiability is thus about the guarantee of this smooth behavior.

Continuity of Functions

Continuity is an essential concept in mathematics, especially when dealing with integrals. A function is deemed continuous if small changes in the input lead to small changes in the output. In the context of our problem, we have a continuous function \(f\), which means there are no breaks or jumps in its graph.
By another part of the Fundamental Theorem of Calculus, the integral \(g(x) = \int_{0}^{x} f(t) \, dt\) of a continuous function \(f\) will be continuous as well. This example illustrates the inherent property of integrals preserving the continuity of the functions they are derived from, ensuring that \(g\) forms a smooth, unbroken line.

Local Extrema

Local extrema refer to the points in a function where it achieves a local maximum or minimum. To identify these in the function \(g\), we normally look for where its derivative \(g'(x)\) is zero — these are critical points. However, since \(g'(x) = f(x) > 0\) at all points, \(g(x)\) can never hit zero. Thus, \(g\) doesn't have any local maxima or minima in its domain.
These points are significant as they can indicate the peaks and valleys on the graph of \(g\). The absence of local extrema here shows a constantly increasing \(g\), given \(f\) is always positive.

Inflection Points

An inflection point occurs where the curve of a function changes concavity, such as from concave up to concave down or vice versa. To find these points in \(g(x)\), we'd need \(g''(x)\), the second derivative, to change signs. This requires \(f'(x)\) to shift from positive to negative or vice versa.
In our scenario, since \(f\) has been stated to have a positive derivative across all \(x\), \(f'(x)\) doesn't change signs. Consequently, \(g(x)\) does not exhibit any inflection points. Understanding inflection points is crucial for analyzing how the curvature of a graph evolves over its path.