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Chapter 4: Problem 13

Sketch the graph of \(f\) $$f(x)=\left(\frac{2}{5}\right)^{-x}$$

Short Answer

Expert verified
The graph of \(f(x) = \left(\frac{5}{2}\right)^{x}\) is an increasing exponential curve with a horizontal asymptote at \(y = 0\).

Step by step solution

01

Recognize the Function Type

The function given is \(f(x) = \left(\frac{2}{5}\right)^{-x}\). This function involves an exponential expression with a negative exponent \(-x\). This is an exponential decay function turned into a growth function due to the negative exponent.
02

Rewrite the Function for Easier Analysis

The function can be re-expressed as \(f(x) = \left(\frac{5}{2}\right)^{x}\). The negative exponent flips the base from \(\frac{2}{5}\) to \(\frac{5}{2}\), changing the expression to exponential growth.
03

Find the Key Points

To sketch the graph, first evaluate key points. For example:- At \(x = 0\), \(f(0) = \left(\frac{5}{2}\right)^{0} = 1\).- At \(x = 1\), \(f(1) = \left(\frac{5}{2}\right)^{1} = \frac{5}{2}\).- At \(x = -1\), \(f(-1) = \left(\frac{5}{2}\right)^{-1} = \frac{2}{5}\).
04

Determine the Asymptote

The horizontal asymptote is \(y = 0\). As \(x\) goes to negative infinity, \(f(x)\) approaches zero but never actually reaches it because the function is an exponential growth from \(-\infty\).
05

Sketch the Graph

Plot the points: \((0,1)\), \((1, \frac{5}{2})\), \((-1, \frac{2}{5})\). Connect these points with a smooth curve that becomes very close to the x-axis as \(x\) decreases and rises sharply as \(x\) increases, illustrating exponential growth.

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Key Concepts

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

Exponential Growth
In mathematics, exponential growth describes a process where quantities increase rapidly at a constant rate relative to their current value. The function given as an example, \( f(x) = \left(\frac{5}{2}\right)^x \), shows exponential growth. This is because the base \( \frac{5}{2} \) is greater than 1, leading to increases as \( x \) becomes more positive.
  • For instance, when \( x = 1 \), \( f(1) = \frac{5}{2} \).
  • When \( x = 2 \), \( f(2) = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \).
This illustrates how quickly the values of the function rise.
The exponential growth factor in this problem is evident when analyzing the rewritten form of the original function. With exponential growth, as \( x \) increases, \( f(x) \) grows without bound.
Graph Sketching
Graph sketching involves plotting key points and understanding the behavior of a function to visually represent it accurately. It includes identifying important attributes such as growth or decay, intercepts, and asymptotes.
  • First, identify key points where the function is evaluated, such as \( (0,1) \), \( (1, \frac{5}{2}) \), and \( (-1, \frac{2}{5}) \).
  • Plot these points on a coordinate plane.
  • The shape of the graph for \( f(x) = \left(\frac{5}{2}\right)^x \) reflects exponential growth. It starts near the x-axis and curves sharply upwards as \( x \) increases.
Understanding the steep curve of the exponential function helps in sketching, as its growth rate results in a graph that moves quickly away from the x-axis.
Exponential Decay
Exponential decay is characterized by a decrease in quantities at a rate proportional to their current value. The initial function \( f(x) = \left(\frac{2}{5}\right)^{-x} \) indicates that the negative exponent turns the traditional decay into growth.
  • This transformation involves converting the base and negating the exponent, transforming \( \frac{2}{5} \) decay to \( \frac{5}{2} \) growth.
  • It is the inverse operation of growth, where as \( x \) goes negative, \( f(x) \) approaches zero.
Initially, it seems like an exponential decay problem, but manipulating the exponent reveals the true growth nature of the function. Typically, decay rates result in graphs that approach an asymptote more gradually than seen in growth.
Asymptotes
Asymptotes are lines that a graph approaches but never touches. For the given function, the key asymptote is the horizontal asymptote at \( y = 0 \). This can be observed when values of \( x \) become very large negatively, where the function seems to "flatten out" near the x-axis.
  • The graph approaches \( y = 0 \) for both decay and growth; however, in exponential growth, it moves away rapidly as \( x \) increases.
  • The asymptotic behavior is a crucial feature that gives insight into the end-behavior of exponential functions.
Identifying asymptotes is essential, as they influence how the graph behaves near extreme values of \( x \) and provide clues about the graph's long-term behavior.

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Most popular questions from this chapter

Exer. \(61-62:\) Graph \(f\) and \(g\) on the same coordinate plane, and estimate the solution of the inequality \(f(x) \geq g(x)\) $$f(x)=x^{3}-3.5 x^{2}+3 x ; \quad g(x)=\log 3 x$$

If a savings fund pays interest at a rate of \(6 \%\) per year compounded semiannully, how much money invested now will amount to \(\$ 5000\) after 1 year?

Approximate \(x\) to three significant figures. (a) \(\log x=1.8965\) (b) \(\log x=4.9680\) (c) \(\log x=-2.2118\) (d) \(\ln x=3.7\) (e) \(\ln x=0.95\) (f) \(\ln x=-5\)

Electron energy The energy \(E(x)\) of an electron after passing through material of thickness \(x\) is given by the equation \(E(x)=E_{0} e^{-x k_{0} y},\) where \(E_{0}\) is the initial energy and \(x_{0}\) is the radiation length.Express, in terms of \(E_{0}\), the energy of an electron after it passes through material of thickness \(x_{0}\) (b) Express, in terms of \(x_{0}\), the thickness at which the electron loses \(99 \%\) of its initial energy.

Sketch the graph of \(f\). $$f(x)=\ln |x|$$
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