Problem 4
Question
If \(f(x)\) is continuous such that \(|f(x)| \leq 1, \forall x \in R\) and \(g(x)=\frac{e^{f(x)}-e^{|f(x)|}}{e^{f(x)}+e^{|f(x)|}}\), then range of \(g(x)\) is (a) \([0,1]\) (b) \(\left[0, \frac{e^{2}+1}{e^{2}-1}\right]\) (c) \(\left[0, \frac{e^{2}-1}{e^{2}+1}\right]\) (d) \(\left[\frac{1-e^{2}}{1+e^{2}}, 0\right]\)
Step-by-Step Solution
Verified Answer
The range of \( g(x) \) is \( \left[ \frac{1-e^{2}}{1+e^{2}}, 0 \right] \).
1Step 1: Understand the range of f(x)
The function \(|f(x)| \leq 1\) suggests that the values of \(f(x)\) are within the interval \([-1, 1]\). This forms the domain over which we need to analyze \(g(x)\).
2Step 2: Substitute boundary values into g(x)
We need to analyze \(g(x)\) for boundary values \(f(x) = -1\) and \(f(x) = 1\) due to their potential to yield extreme function values. For \(f(x) = 1\), we get \(g(1) = \frac{e^{1} - e^{1}}{e^{1} + e^{1}} = 0\). For \(f(x) = -1\), compute \(g(-1) = \frac{e^{-1} - e^{1}}{e^{-1} + e^{1}} = \frac{1/e - e}{1/e + e}\).
3Step 3: Simplify g(-1) for the range
Calculate \(g(-1) = \frac{1/e - e}{1/e + e} = \frac{1 - e^2}{1 + e^2}\). This value represents the lower bound of \(g(x)\), while the upper bound is defined by \(g(1)=0\). Hence, the range of \(g(x)\) includes values between \(\frac{1 - e^2}{1 + e^2}\) and \(0\).
4Step 4: Analyze the behavior of exponential terms
Since \(|f(x)| \leq 1\), \(f(x)\) and \(|f(x)|\) can both be -1, 0, or 1. The function \(g(x)\) effectively calculates the limit behavior: at \(f(x) = 0, g(x)\) results in 0; when \(f(x)\) swings to its extremes, \(-1 \) or \(1\), the function takes on values delimiting \(\frac{1 - e^2}{1 + e^2}\) to 0.
Key Concepts
Range of FunctionsExponential FunctionsBoundaries of Functions
Range of Functions
The range of a function describes all possible output values it can produce, often seen as the "y-values" on a graph. To determine this range, we explore how the function behaves across its domain.
Understanding ranges is a crucial aspect of working with functions, as it informs us of all possible outputs and thus, potential solutions.
- In our exercise, the function is given as \(g(x)\).
- The task was to identify the range of \(g(x)\), knowing it depends on \(f(x)\) being continuous and bounded by \([-1, 1]\).
- We substitute extreme values into \(g(x)\) to find its upper and lower bounds.
Understanding ranges is a crucial aspect of working with functions, as it informs us of all possible outputs and thus, potential solutions.
Exponential Functions
Exponential functions are characterized by their constant growth rate, which is what makes them so intriguing. They have the general form \(a^{x}\), where the base \(a\) is a constant and the exponent \(x\) is the variable.
This highlights an essential aspect of exponential functions: their potential to magnify small changes in \(f(x)\) because of their inherent nature to grow rapidly.
- In the exercise, exponential functions \(e^{f(x)}\) and \(e^{|f(x)|}\) were used to derive \(g(x)\).
- The exponential nature of these functions is central to how they transform values from \(f(x)\) into the numerator and denominator of \(g(x)\).
This highlights an essential aspect of exponential functions: their potential to magnify small changes in \(f(x)\) because of their inherent nature to grow rapidly.
Boundaries of Functions
The boundaries of a function are crucial in understanding its behavior, especially when analyzed for continuity and absolute limits.
Thus, by understanding the constraints on \(f(x)\), we make more informed assessments of the boundaries and eventual range of \(g(x)\). These boundaries not only illustrate the limits but also guide us through the subtle variations within the domain, thereby defining the scope of the function's output.
- Our exercise discusses the boundaries of \(f(x)\) given by \(|f(x)| \leq 1\).
- This indicates that \(f(x)\) stays within \([-1, 1]\), directing our analysis of \(g(x)\).
Thus, by understanding the constraints on \(f(x)\), we make more informed assessments of the boundaries and eventual range of \(g(x)\). These boundaries not only illustrate the limits but also guide us through the subtle variations within the domain, thereby defining the scope of the function's output.
Other exercises in this chapter
Problem 2
Let \(f: X \rightarrow Y, f(x)=\sin x+\cos x+2 \sqrt{2}\) is invertible, then \(X \rightarrow Y\) is/are (a) \(\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right] \righ
View solution Problem 3
The range of values of \(a\) so that all the roots of the equation \(2 x^{3}-3 x^{2}-12 x+a=0\) are real and distinct belongs to (a) \((7,20)\) (b) \((-7,20)\)
View solution Problem 5
Let \(f(x)=\sqrt{|x|-\\{x\\}}\) (where \\{\\} denotes the fractional part of \(x\) and \(X, Y\) and its domain and range respectively, then (a) \(f: X \rightarr
View solution Problem 6
If the graphs of the functions \(y=\ln x\) and \(y=a x\) intersect at exactly two points, then \(a\) must be (a) \((0, e)\) (b) \(\left(\frac{1}{e}, 0\right)\)
View solution