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Chapter 4: Problem 48

Let \(f\) be a function with \(f(x)>0\) for all \(x .\) Set \(g=1 / f\) (a) If \(f\) is increasing in an interval around \(x_{0},\) what about \(g ?\) (b) If \(f\) has a local maximum at \(x_{1},\) what about \(g ?\) (c) If \(f\) is concave down at \(x_{2},\) what about \(g ?\)

Short Answer

Expert verified
(a) g is decreasing. (b) g has a local minimum. (c) g is concave up.

Step by step solution

01

Understanding positivity of f(x)

Given that \( f(x) > 0 \) for all \( x \), this means \( f \) is always positive over its domain, which ensures \( g(x) = \frac{1}{f(x)} \) is well-defined since the denominator is never zero.
02

Analyzing g for increasing f

If \( f \) is increasing around \( x_0 \), then the derivative \( f'(x) > 0 \) around \( x_0 \). The function \( g(x) = \frac{1}{f(x)} \) is decreasing in this interval because as \( f(x) \) increases, \( g(x) \) decreases.
03

Deriving implications of a local maximum for f

If \( f \) has a local maximum at \( x_1 \), then \( f'(x_1) = 0 \) and \( f'(x) \) changes from positive to negative as \( x \) passes through \( x_1 \). Consequently, \( g(x) \) has a local minimum at \( x_1 \) because \( g(x) = \frac{1}{f(x)} \) and the derivative of \( g \) changes from negative to positive.
04

Evaluating concavity of g from concavity of f

If \( f \) is concave down at \( x_2 \), then \( f''(x_2) < 0 \). For \( g(x) = \frac{1}{f(x)} \), the second derivative \( g''(x) \) depends on \( f''(x) \) and is positive at \( x_2 \), indicating that \( g \) is concave up in this region.

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Key Concepts

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

Increasing Functions
When we talk about functions that are increasing, it means as the input (or x-value) goes up, the output (or y-value) of the function goes up as well.
In mathematical terms, for a function \(f\), if \(f'(x) > 0\) for all \(x\) in an interval, the function is increasing in that interval.
For example, consider a hill you are walking up. As you take a step forward, you go higher. This is similar to how an increasing function operates.
  • If \(f(x)\) increases, then the rate of change or slope of \(f\) is positive.
  • In the context of the given function \(g(x) = \frac{1}{f(x)}\), if \(f\) is increasing, \(g\) will be decreasing because the reciprocal of a larger number becomes smaller.
This inverse relationship is fundamental when dealing with reciprocal functions.
Local Extrema
Local extrema refer to the "peaks" and "valleys" of a function within a specific period or interval.
They are important because they tell us where a function changes its increasing or decreasing behavior.
There are two types of local extrema:
  • Local Maximum: The highest point in a particular section of the graph, similar to standing at the top of a hill.
  • Local Minimum: The lowest point, akin to being at the bottom of a valley.
For the function \(f\), if there is a local maximum at \(x_1\), this implies \(f'(x_1)=0\), and the derivative changes sign from positive to negative.
In the reciprocal function \(g(x) = \frac{1}{f(x)}\), such a point will actually be a local minimum because as \(f\) decreases beyond its peak, \(g\) begins increasing, behaving like the opposite of \(f\).
Concavity
Concavity tells us about the "curved shape" of the graph of a function. Specifically, it informs whether the function is "cupping" upward or downward.
  • If a function is concave up, it looks like a cup (or a smile). Here, the second derivative \(f''(x) > 0\).
  • If a function is concave down, it resembles an upside-down cup (or a frown), and \(f''(x) < 0\).
In the problem, \(f\) is concave down at \(x_2\) which means \(f''(x_2) < 0\).
Thus, when considering the function \(g(x) = \frac{1}{f(x)}\), it turns out to be concave up at \(x_2\).
This happens because the reciprocal function \(g\) takes on the opposite concavity of \(f\). Understanding this helps to predict the behavior and shape of functions under transformations such as reciprocal.

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Most popular questions from this chapter

Give an example of: A member of the family \(f(x)=a x^{3}-b x\) that has no critical points.

Let \(g(x)=x-k e^{x},\) where \(k\) is any constant. For what value(s) of \(k\) does the function \(g\) have a critical point?

In a romantic relationship between Angela and Brian, who are unsuited for each other, \(a(t)\) represents the affection Angela has for Brian at time \(t\) days after they meet, while \(b(t)\) represents the affection Brian has for Angela at time \(t .\) If \(a(t)>0\) then Angela likes Brian; if \(a(t)<0\) then Angela dislikes Brian; if \(a(t)=0\) then Angela neither likes nor dislikes Brian. Their affection for each other is given by the relation \((a(t))^{2}+(b(t))^{2}=c\) where \(c\) is a constant. (a) Show that \(a(t) \cdot a^{\prime}(t)=-b(t) \cdot b^{\prime}(t)\) (b) At any time during their relationship, the rate per day at which Brian's affection for Angela changes is \(b^{\prime}(t)=-a(t) .\) Explain what this means if Angela (i) Likes Brian, (ii) Dislikes Brian. (c) Use parts (a) and (b) to show that \(a^{\prime}(t)=b(t) .\) Explain what this means if Brian (i) Likes Angela, (ii) Dislikes Angela. (d) If \(a(0)=1\) and \(b(0)=1\) who first dislikes the other?

For positive constants \(A\) and \(B,\) the force, \(F,\) between two atoms in a molecule at a distance \(r\) apart is given by $$F=-\frac{A}{r^{2}}+\frac{B}{r^{3}}$$ (a) How fast does force change as \(r\) increases? What type of units does it have? (b) If at some time \(t\) the distance is changing at a rate \(k\) at what rate is the force changing with time? What type of units does this rate of change have?

Are the statements in Problems true or false for a function \(f\) whose domain is all real numbers? If a statement is true,explain how you know. If a statement is false, give a counterexample. If \(f^{\prime \prime}(x)\) is continuous and the graph of \(f\) has an inflection point at \(x=p,\) then \(f^{\prime \prime}(p)=0\).
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