Given The Function F ( X ) = 3 ( 1 2 ) X F(x) = 3\left(\frac{1}{2}\right)^x F ( X ) = 3 ( 2 1 ​ ) X , Complete The Table And Determine The Characteristics Of The Function.$[ \begin{array}{|c|c|} \hline x & Y \ \hline -2 & 3 \times \left(\frac{1}{2}\right)^{-2} \ \hline -1 & 3

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The given function is f(x)=3(12)xf(x) = 3\left(\frac1}{2}\right)^x3(21)x. This is an exponential function with a base of 12\frac{1{2} and a coefficient of 33. The function is defined for all real values of xx.

Completing the Table

To complete the table, we need to calculate the values of yy for different values of xx. We will use the given function to calculate the values of yy.

x y
-2 3 × (1/2)^(-2)
-1 3
0 3 × (1/2)^0
1 3 × (1/2)^1
2 3 × (1/2)^2

Let's calculate the values of yy for each value of xx.

Calculating y for x = -2

To calculate the value of yy for x=2x = -2, we substitute x=2x = -2 into the function:

f(2)=3(12)2f(-2) = 3\left(\frac{1}{2}\right)^{-2}

Using the property of exponents that an=1ana^{-n} = \frac{1}{a^n}, we can rewrite the expression as:

f(2)=3(21)2f(-2) = 3\left(\frac{2}{1}\right)^2

Simplifying the expression, we get:

f(2)=3×4=12f(-2) = 3 \times 4 = 12

So, the value of yy for x=2x = -2 is 1212.

Calculating y for x = -1

To calculate the value of yy for x=1x = -1, we substitute x=1x = -1 into the function:

f(1)=3(12)1f(-1) = 3\left(\frac{1}{2}\right)^{-1}

Using the property of exponents that an=1ana^{-n} = \frac{1}{a^n}, we can rewrite the expression as:

f(1)=3(21)f(-1) = 3\left(\frac{2}{1}\right)

Simplifying the expression, we get:

f(1)=3×2=6f(-1) = 3 \times 2 = 6

So, the value of yy for x=1x = -1 is 66.

Calculating y for x = 0

To calculate the value of yy for x=0x = 0, we substitute x=0x = 0 into the function:

f(0)=3(12)0f(0) = 3\left(\frac{1}{2}\right)^0

Using the property of exponents that a0=1a^0 = 1, we can rewrite the expression as:

f(0)=3×1=3f(0) = 3 \times 1 = 3

So, the value of yy for x=0x = 0 is 33.

Calculating y for x = 1

To calculate the value of yy for x=1x = 1, we substitute x=1x = 1 into the function:

f(1)=3(12)1f(1) = 3\left(\frac{1}{2}\right)^1

Simplifying the expression, we get:

f(1)=3×12=32f(1) = 3 \times \frac{1}{2} = \frac{3}{2}

So, the value of yy for x=1x = 1 is 32\frac{3}{2}.

Calculating y for x = 2

To calculate the value of yy for x=2x = 2, we substitute x=2x = 2 into the function:

f(2)=3(12)2f(2) = 3\left(\frac{1}{2}\right)^2

Simplifying the expression, we get:

f(2)=3×14=34f(2) = 3 \times \frac{1}{4} = \frac{3}{4}

So, the value of yy for x=2x = 2 is 34\frac{3}{4}.

Updated Table

x y
-2 12
-1 6
0 3
1 1.5
2 0.75

Characteristics of the Function

The function f(x)=3(12)xf(x) = 3\left(\frac{1}{2}\right)^x is an exponential function with a base of 12\frac{1}{2} and a coefficient of 33. The function is defined for all real values of xx.

Domain of the Function

The domain of the function is all real numbers, i.e., x(,)x \in (-\infty, \infty).

Range of the Function

The range of the function is all positive real numbers, i.e., y(0,)y \in (0, \infty).

End Behavior of the Function

As xx approaches negative infinity, yy approaches infinity.

As xx approaches positive infinity, yy approaches 00.

Intercepts of the Function

The function has a yy-intercept at (0,3)(0, 3).

The function has no xx-intercepts.

Asymptotes of the Function

The function has a horizontal asymptote at y=0y = 0.

The function has no vertical asymptotes.

Increasing and Decreasing Intervals

The function is decreasing on the interval (,)(-\infty, \infty).

Local Maxima and Minima

The function has no local maxima or minima.

Concavity of the Function

The function is concave down on the interval (,)(-\infty, \infty).

Inflection Points

The function has no inflection points.

Second Derivative

The second derivative of the function is:

f(x)=3(12)xln(12)2f''(x) = 3\left(\frac{1}{2}\right)^x \ln\left(\frac{1}{2}\right)^2

In the previous article, we explored the function f(x)=3(12)xf(x) = 3\left(\frac{1}{2}\right)^x and completed a table to determine its characteristics. In this article, we will answer some frequently asked questions about the function and its characteristics.

Q: What is the domain of the function?

A: The domain of the function is all real numbers, i.e., x(,)x \in (-\infty, \infty).

Q: What is the range of the function?

A: The range of the function is all positive real numbers, i.e., y(0,)y \in (0, \infty).

Q: How does the function behave as x approaches negative infinity?

A: As xx approaches negative infinity, yy approaches infinity.

Q: How does the function behave as x approaches positive infinity?

A: As xx approaches positive infinity, yy approaches 00.

Q: What is the y-intercept of the function?

A: The function has a yy-intercept at (0,3)(0, 3).

Q: Does the function have any x-intercepts?

A: No, the function has no xx-intercepts.

Q: Does the function have any vertical asymptotes?

A: No, the function has no vertical asymptotes.

Q: Is the function increasing or decreasing?

A: The function is decreasing on the interval (,)(-\infty, \infty).

Q: Does the function have any local maxima or minima?

A: No, the function has no local maxima or minima.

Q: Is the function concave up or concave down?

A: The function is concave down on the interval (,)(-\infty, \infty).

Q: Does the function have any inflection points?

A: No, the function has no inflection points.

Q: How do you calculate the second derivative of the function?

A: The second derivative of the function is:

f(x)=3(12)xln(12)2f''(x) = 3\left(\frac{1}{2}\right)^x \ln\left(\frac{1}{2}\right)^2

Q: What does the second derivative tell us about the function?

A: The second derivative is negative for all values of xx, indicating that the function is concave down on the interval (,)(-\infty, \infty).

Q: How can you use the function to model real-world situations?

A: The function can be used to model situations where a quantity decreases exponentially over time. For example, the function can be used to model the decay of a radioactive substance or the decrease in population of a species over time.

Q: How can you graph the function?

A: The function can be graphed using a graphing calculator or a computer algebra system. The graph of the function will be a decreasing exponential curve that approaches the x-axis as x approaches positive infinity.

Q: What are some common applications of the function?

A: The function has many applications in fields such as physics, engineering, and economics. Some common applications of the function include:

  • Modeling the decay of radioactive substances
  • Modeling the decrease in population of a species over time
  • Modeling the growth of a population over time
  • Modeling the decay of a chemical reaction
  • Modeling the growth of a financial investment over time

We hope this Q&A article has helped you understand the function and its characteristics. If you have any further questions, please don't hesitate to ask.