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Determine the end behavior of the following transcendental functions by analyzing appropriate limits. Then provide a simple sketch of the associated graph, showing asymptotes if they exist. $$f(x)=\frac{50}{e^{2 x}}$$

Short Answer

Expert verified
Answer: As x approaches both positive and negative infinity, the function approaches 0, with a horizontal asymptote at y = 0.

Step by step solution

01

Analyze the limit as x approaches positive and negative infinity

First, let's find the limit as x approaches positive infinity: $$\lim_{x \to \infty} \frac{50}{e^{2x}}$$ Since the exponential function grows faster than any polynomial function, this limit will go to 0: $$\lim_{x \to \infty} \frac{50}{e^{2x}} = 0$$ Now, let's find the limit as x approaches negative infinity: $$\lim_{x \to -\infty} \frac{50}{e^{2x}}$$ Since \(e^x\) will go to infinity as x approaches negative infinity, the function will also go to 0: $$\lim_{x \to -\infty} \frac{50}{e^{2x}} = 0$$ This means, as x approaches both positive and negative infinity, the function approaches 0.
02

Determine any horizontal or vertical asymptotes

Since the limit of the function as x approaches both positive and negative infinity is 0, there exists a horizontal asymptote at y=0. There are no vertical asymptotes in this function as the denominator will never equal 0.
03

Sketch the graph of the function

Based on the information gathered in the previous steps, we will now sketch the graph of the function $$f(x)=\frac{50}{e^{2 x}}$$ 1. The function approaches 0 as x approaches both positive and negative infinity, forming a horizontal asymptote at y = 0. 2. There are no vertical asymptotes. 3. The function starts at a positive value and then decreases as x increases. Using this information, we can sketch a simple graph of the function. Note that the sketch may not be to scale, but it serves to demonstrate the overall behavior of the function.

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Key Concepts

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

End Behavior
End behavior refers to how a function behaves as the input, or x-value, approaches extreme values such as positive or negative infinity. In the context of transcendental functions like \( f(x) = \frac{50}{e^{2x}} \), understanding the end behavior helps in predicting how the function will trend in the long term.

For our function, as \( x \) approaches positive infinity, the term \( e^{2x} \) grows extremely large at an exponential rate. Thus, the denominator becomes very large, and the whole fraction approaches zero. This illustrates the rapid decay or shrinking of \( f(x) \) towards zero.
  • The limit \( \lim_{x \to \infty} \frac{50}{e^{2x}} = 0 \) clearly shows the end behavior as x grows larger.
  • Conversely, as \( x \) approaches negative infinity, \( e^{2x} \) again tends to grow towards positive infinity, due to the negative sign flipping in the exponent.
  • Therefore, \( \lim_{x \to -\infty} \frac{50}{e^{2x}} = 0 \) is another indication of decay.
This pattern of behavior defines the function's trend as reaching toward zero, exemplifying classic behaviors of functions involving exponential components, contributing to understanding and sketching its long-term graph patterns.
Limits
Limits are a fundamental concept in calculus, used to find the value that a function approaches as the input approaches a specific point. For transcendental functions, this concept is crucial in determining end behavior and asymptotes.

In analyzing \( f(x) = \frac{50}{e^{2x}} \), calculating the limit as \( x \to \infty \) demonstrates how \( f(x) \) behaves as x grows without bounds.
  • The limit \( \lim_{x \to \infty} \frac{50}{e^{2x}} \) yields a result of 0, indicating the output value diminishes.
  • Similarly, for \( x \to -\infty \), the function also approaches 0, \( \lim_{x \to -\infty} \frac{50}{e^{2x}} = 0 \).
Understanding these limits gives insight into how the function behaves overall. Limits offer precise information on the approach direction of \( f(x) \) as \( x \) becomes exceedingly large or small, vital for examining its continuity and endpoints.
Horizontal Asymptotes
Horizontal asymptotes graphically represent a line that a function approaches but never touches. These are commonly found by analyzing limits as \( x \) tends towards infinity or negative infinity.

For the function \( f(x) = \frac{50}{e^{2x}} \), the previously calculated limits both approaching 0 imply the presence of a horizontal asymptote along the line \( y = 0 \).
  • Since as \( x \to \infty \) and \( x \to -\infty \), \( f(x) \to 0 \), this establishes \( y = 0 \) as a horizontal asymptote.
  • There are no vertical asymptotes because the exponential expression in the denominator never reaches zero; hence the function remains continuous.
Recognizing horizontal asymptotes is fundamental in graph sketches, as they guide the function’s trend at the extremes of the x-axis. They provide students with a visual cue for understanding the behavior of the function outside of center values.

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Most popular questions from this chapter

Evaluate \(\lim _{x \rightarrow \infty} f(x)\) \(f(x)\) and \(\lim _{x \rightarrow-\infty} f(x)\) Then state the horizontal asymptote(s) of \(f\). Confirm your findings by plotting \(f\) $$f(x)=\frac{3 e^{x}+e^{-x}}{e^{x}+e^{-x}}$$

The limit at infinity \(\lim _{x \rightarrow \infty} f(x)=L\) means that for any \(\varepsilon>0\) there exists \(N>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad x>N.$$ Use this definition to prove the following statements. $$\lim _{x \rightarrow \infty} \frac{10}{x}=0$$

a. Given the graph of \(f\) in the following figures, find the slope of the secant line that passes through (0,0) and \((h, f(h))\) in terms of \(h\) for \(h>0\) and \(h<0\) b. Analyze the limit of the slope of the secant line found in part (a) as \(h \rightarrow 0^{+}\) and \(h \rightarrow 0^{-} .\) What does this tell you about the line tangent to the curve at (0,0)\(?\) $$f(x)=x^{1 / 3}$$

Assume you invest 250 dollars at the end of each year for 10 years at an annual interest rate of \(r\) The amount of money in your account after 10 years is \(A=\frac{250\left((1+r)^{10}-1\right)}{r} .\) Assume your goal is to have 3500 dollars in your account after 10 years. a. Use the Intermediate Value Theorem to show that there is an interest rate \(r\) in the interval \((0.01,0.10)-\) between \(1 \%\) and \(10 \%-\) that allows you to reach your financial goal. b. Use a calculator to estimate the interest rate required to reach your financial goal.

Use the following definitions. Assume fexists for all \(x\) near a with \(x>\) a. We say the limit of \(f(x)\) as \(x\) approaches a from the right of a is \(L\) and write \(\lim _{x \rightarrow a^{+}} f(x)=L,\) if for any \(\varepsilon>0\) there exists \(\delta>0\) such that $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0< x-a<\delta$$ Assume fexists for all \(x\) near a with \(x < \) a. We say the limit of \(f(x)\) as \(x\) approaches a from the left of a is \(L\) and write \(\lim _{x \rightarrow a^{-}} f(x)=L,\) if for any \(\varepsilon > 0 \) there exists \(\delta > 0\) such that $$|f(x)-L| < \varepsilon \quad \text { whenever } \quad 0< a-x <\delta$$ Why is the last inequality in the definition of \(\lim _{x \rightarrow a} f(x)=L,\) namely, \(0<|x-a|<\delta,\) replaced with \(0

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