Determining the Range of a Specific Real-Valued Function

In this article, we will explore a specific real-valued function defined by the equation 2fsin(x) - fcos(x) x, where the domain of the function f is [1, -1]. Our goal is to determine the range of this function. We will approach this problem by substituting specific values of x, solving the system of equations, and analyzing the continuity and nature of the function.

Introduction

The given functional equation is:

2fsin(x) - fcos(x) x

Where f is a real-valued function and x is within the interval [1, -1]. To find the range of f, we will follow a step-by-step approach to determine its values at key points within the domain and then deduce the range based on these values and the continuity of the function.

Step 1: Analyzing the Equation

First, we will substitute specific values to gain insight into the function f.

Substitution

x 0:

2fsin(0) - fcos(0) 0

This simplifies to:

2f0 - f1 0 (1)

x frac{pi}{2}:

2fsin(frac{pi}{2}) - fcos(frac{pi}{2}) frac{pi}{2}

This simplifies to:

2f1 - f0 frac{pi}{2} (2)

x frac{pi}{4}:

2fsin(frac{pi}{4}) - fcos(frac{pi}{4}) frac{pi}{4}

This simplifies to:

2f(frac{sqrt{2}}{2}) - f(frac{sqrt{2}}{2}) frac{pi}{4}

Therefore:

3f(frac{sqrt{2}}{2}) frac{pi}{4} (3)

Step 2: Solving the System of Equations

From equations (1) and (2), we have a system of equations involving f0 and f1.

From (1):

f1 -2f0

Substituting f1 into (2):

2-2f0 - f0 frac{pi}{2}

Which simplifies to:

-4f0 - f0 frac{pi}{2}

-3f0 frac{pi}{2}

Therefore:

f0 -frac{pi}{6}

Substituting f0 back into equation (1):

f1 -2(-frac{pi}{6}) frac{pi}{3}

Step 3: Finding the Range of f

So far, we have determined the following:

f(0) -frac{pi}{6}

f(1) frac{pi}{3}

f(frac{sqrt{2}}{2}) frac{pi}{12}

To check for continuity and determine the range of f, we note that the equation 2fsin(x) - fcos(x) x suggests a continuous relationship between the inputs and outputs of f. Since f is defined at three points and appears continuous over the interval [1, -1], the range of f is likely to be the interval between the minimum and maximum values computed.

Step 4: Conclusion

Given that f is determined at three points and f is likely continuous, the range of f can be inferred as:

Range of f [-frac{pi}{6}, frac{pi}{3}]

Thus, the range of f is:

boxed{[-frac{pi}{6}, frac{pi}{3}]}

Key Takeaways

Substituting specific values into the functional equation can provide insight into the behavior of the function. Solving the resulting system of equations can help determine the values of the function at key points. Continuity of the function can be used to determine the range based on the values at key points and the behavior of the function in between.

Keywords

Functional Equation Real-Valued Function Range Determination