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The function \(f\) is increasing on the interval \(I\). Prove: (a) if \(g(x)=-f(x)\), then \(g\) is decreasing on \(I ;(b)\) if \(h(x)=1 / f(x)\), and \(f(x)>0\) on \(I\), then \(h\) is decreasing on \(I\).

Short Answer

Expert verified
If \(g(x) = -f(x)\), then \(g\) is decreasing. If \(h(x) = \frac{1}{f(x)}\) and \(f(x) > 0\), then \(h\) is decreasing.

Step by step solution

01

- Understand the Given Information

Given that the function \(f\) is increasing on the interval \(I\). This means that for any \(x_1, x_2 \in I\) with \(x_1 < x_2\), we have \(f(x_1) < f(x_2)\).
02

- Define the Function for (a)

Let \(g(x) = -f(x)\). We need to show that \(g\) is decreasing on the interval \(I\).
03

- Show for (a) that \(g\) is Decreasing

Consider any two points \(x_1, x_2 \in I\) such that \(x_1 < x_2\). Since \(f\) is increasing, \(f(x_1) < f(x_2)\). Thus, \(-f(x_1) > -f(x_2)\), which implies that \(g(x_1) > g(x_2)\). Therefore, \(g\) is decreasing on \(I\).
04

- Define the Function for (b)

Let \(h(x) = \frac{1}{f(x)}\). We need to show that \(h\) is decreasing on the interval \(I\) given that \(f(x) > 0\) for all \(x \in I\).
05

- Show for (b) that \(h\) is Decreasing

Consider any two points \(x_1, x_2 \in I\) such that \(x_1 < x_2\). Since \(f\) is increasing and \(f(x) > 0\), we have \(f(x_1) < f(x_2)\). Taking the reciprocal, we get \(\frac{1}{f(x_1)} > \frac{1}{f(x_2)}\). Thus, \(h(x_1) > h(x_2)\), which means \(h\) is decreasing on \(I\).

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Key Concepts

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

Monotonic Functions
Monotonic functions are a fundamental concept in mathematics. A function is called monotonic if it is either entirely non-increasing or non-decreasing on a given interval. This implies that the function either consistently increases or consistently decreases.

When we say a function 饾憮(x) is increasing on an interval I, it means that for any two points x1 and x2 in I where x1 < x2, the function value at x1 is less than the function value at x2, i.e., 饾憮(x1) < 饾憮(虫2). Conversely, a function 饾憮(x) is decreasing on an interval I if for any two points x1 and x2 in I where x1 < x2, the function value at x1 is greater than the function value at x2, i.e., 饾憮(x1) > 饾憮(虫2).

Monotonic functions are important because they help us understand the behavior of functions over intervals. They can also help us determine if a function has a unique solution crossing a specific value.
Reciprocal Functions
Reciprocal functions involve the inverse of another function. For a given function 饾憮(x), its reciprocal function is defined as h(x) = 1/饾憮(x). This relationship introduces several interesting properties, especially when dealing with monotonic functions.

In the given exercise, we saw that if 饾憮(x) is an increasing function and 饾憮(x) > 0, then the reciprocal function h(x) = 1/饾憮(x) is a decreasing function. This can be understood by considering the nature of reciprocals. Since 饾憮(x) is increasing, any increase in the value of 饾憮(x) will lead to a decrease in the value of 1/饾憮(x) because the denominator is getting larger. Therefore, as 饾憮(x) increases, h(x) = 1/饾憮(x) decreases, showing the reciprocal function's decreasing nature on the interval where 饾憮(x) > 0.
Negative Functions
Negative functions involve multiplying the given function by -1. If 饾憮(x) is given, its negative function can be expressed as 饾憯(x) = -饾憮(x). This transformation impacts the function's monotonicity.

From the exercise, we learned that if 饾憮(x) is increasing, then 饾憯(x) = -饾憮(x) will be decreasing. This is because multiplying a function by -1 reverses its direction. For any two points x1 and x2 in the interval I where x1 < x2, if 饾憮(x1) < 饾憮(x2), then -饾憮(x1) > -饾憮(虫2). This shows that while 饾憮(x) climbs upwards, -饾憮(x) slopes downwards. Negative functions provide a straightforward way to convert increasing functions into decreasing ones, helping us understand the symmetrical nature of function transformations.

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