What exponential function is represented by the graph, f(x) = 2 X (3^x)

1 answer:

3 0

Evaluate 3x at x = –2, –1, 0, 1, and 2.To find the answer, I need to plug in the given values for x, and simplify:Given f(x) = 3–x, evaluate f(–2), f(–1), f(0), f(1), and f(2).To find the answer, I need to plug in the given values for x, and simplify:

Take another look at the values I came up with: they were precisely reversed between the two T-charts. Remember that negative exponents mean that you have to flip the base to the other side of the fraction line. This means that 3–x may also be written as ( 1/3 )x, by taking the "minus" in the exponent and using it to flip the base "3". With this in mind, you should be able to predict the values for the following problem: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

Given g(x) = ( 1/3 )x, evaluate for x = –2, –1, 0, 1, and 2. Plug in the given values for x, and simplify:

This exercise points out two things. First, you really do need to be good with exponents in order to do exponentials (so review the topic, if necessary), and, second, exponential decay (getting smaller and smaller by half (or a third, or...) at each step) is just like exponential growth, except that either the exponent is "negative" (the "–x" in "3–x") or else the base is between 0 and 1 (the "1/3" in "( 1/3 )x").

It will likely be necessary for you to be able to just look at an equation or an expression or a graph and correctly identify which type of change it represents, growth or decay, so go back and study the above examples, if you're not sure of what is going on here.

To be thorough, since 3x models growth: 
...and since 3–x and ( 1/3 )x model decay:

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3 years ago

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Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Otrada [13]

I guess the "5" is supposed to represent the integral sign?

$I=\displaystyle\int_1^4\ln t\,\mathrm dt$

With $n=10$ subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

$\ell_i=1+\dfrac{3(i-1)}{10}$

and right endpoints are given by

$r_i=1+\dfrac{3i}{10}$

where $1\le i\le10$ .

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, $\dfrac{4-1}{10}=\dfrac3{10}$ , and "bases" equal to the values of $\ln t$ at both endpoints of each subinterval. The area of the trapezoid over the $i$ -th subinterval is

$\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)$

Then the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}$

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of $\ln t$ at the average of the subinterval's endpoints, $\dfrac{\ell_i+r_i}2$ . The area of the rectangle over the $i$ -th subinterval is then

$\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}$

so the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}$

c. For Simpson's rule, we find a quadratic interpolation of $\ln t$ over each subinterval given by

$P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}$