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Chapter 3: Problem 29

Use a graphing utility to graph the exponential function. $$y=2^{-x^{2}}$$

Short Answer

Expert verified
The graph of the function \(y = 2^{-x^{2}}\) starts and ends at \(y=0\) for \(x 鈫� 卤鈭瀄). It reaches maximum at \(x=0\) where \(y = 2^{-0^{2}} = 1\). It shows that for larger magnitude of \(x\), \(y\) becomes very close to zero.

Step by step solution

01

Understand the nature of the function

Firstly, note that the function is an exponential function in the form of \(y = a^{f(x)}\), where \(a > 0, a 鈮� 1\), and \(f(x) = -x^2\). This function will always yield positive values since any real number raised to an even power is always positive.
02

Evaluate the function at some points

Calculate the value of \(y\) at a few points as follows. If \(x=0\), \(y = 2^{-0^{2}} = 2^{0} = 1\). For other values, as \(x\) moves away from zero in positive or negative direction, \(y\) should get smaller since we are raising 2 to a negative power which gets bigger in magnitude. So as \(x\) gets larger (either positive or negative), \(y\) approaches zero.
03

Plot the function using a graphing utility

Use a graphing utility such as Desmos or GeoGebra to plot the function \(y = 2^{-x^{2}}\). The curve will rise from \(y=0\) to \(y=1\) at \(x=0\) and then approach back towards \(y=0\) as \(x\) moves from zero.

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Most popular questions from this chapter

A cell site is a site where electronic communications equipment is placed in a cellular network for the use of mobile phones. The numbers \(y\) of cell sites from 1985 through 2008 can be modeled by \(y=\frac{237,101}{1+1950 e^{-0.355 t}}\) where \(t\) represents the year, with \(t=5\) corresponding to \(1985 .\) (Source: CTIA-The Wireless Association) (a) Use the model to find the numbers of cell sites in the years 1985,2000 , and 2006 . (b) Use a graphing utility to graph the function. (c) Use the graph to determine the year in which the number of cell sites will reach 235,000 . (d) Confirm your answer to part (c) algebraically.

The populations \(P\) (in thousands) of Reno, Nevada from 2000 through 2007 can be modeled by \(P=346.8 e^{k t},\) where \(t\) represents the year, with \(t=0\) corresponding to 2000 . In \(2005,\) the population of Reno was about 395,000 . (Source: U.S. Census Bureau) (a) Find the value of \(k\). Is the population increasing or decreasing? Explain. (b) Use the model to find the populations of Reno in 2010 and 2015 . Are the results reasonable? Explain. (c) According to the model, during what year will the population reach \(500,000 ?\)

Use the acidity model given by \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right],\) where acidity \((\mathrm{pH})\) is a measure of the hydrogen ion concentration \(\left[\mathrm{H}^{+}\right]\) (measured in moles of hydrogen per liter) of a solution. Compute \(\left[\mathrm{H}^{+}\right]\) for a solution in which \(\mathrm{pH}=5.8\).

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$10-4 \ln (x-2)=0$$

A sport utility vehicle that costs $$\$ 23,300$$ new has a book value of $$\$ 12,500$$ after 2 years. (a) Find the linear model \(V=m t+b\). (b) Find the exponential model \(V=a e^{k t}\) (c) Use a graphing utility to graph the two models in the same viewing window. Which model depreciates faster in the first 2 years? (d) Find the book values of the vehicle after 1 year and after 3 years using each model. (e) Explain the advantages and disadvantages of using each model to a buyer and a seller.
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