Submit Test: 24-25 Pre-College MathGiven Functions:${ F(x) = 2^x + 1 }$ { G(x) = \left(\frac{1}{3}\right)^x - 2 \} For Each Statement Below, Drag Your TRUE Or FALSE Response To The Appropriate Box.- True- FalseDrag And Drop An

Mar 12, 2025 by ADMIN 229 views

Submit Test: 24-25 Pre-College Math

Problem 1: Functions and Graphs

Given the functions $f(x) = 2^x + 1$ and $g(x) = \left(\frac{1}{3}\right)^x - 2$ , we need to evaluate the following statements and drag our TRUE or FALSE response to the appropriate box.

Statement 1: $f(x) = g(x)$ for all values of $x$

To determine if this statement is true or false, we need to compare the two functions. We can start by setting the two functions equal to each other:

2^x + 1 = \left(\frac{1}{3}\right)^x - 2

However, this equation is not true for all values of $x$ . In fact, the two functions are not equal for any value of $x$ . This is because the base of the exponential function on the left-hand side is 2, while the base of the exponential function on the right-hand side is $\frac{1}{3}$ . These two bases are not equal, so the two functions cannot be equal for all values of $x$ .

Therefore, the statement $f(x) = g(x)$ for all values of $x$ is FALSE.

Statement 2: $f(x) = g(x)$ for $x = 0$

To determine if this statement is true or false, we need to evaluate the two functions at $x = 0$ .

f(0) = 2^0 + 1 = 2

g(0) = \left(\frac{1}{3}\right)^0 - 2 = 1 - 2 = -1

Since $f(0) \neq g(0)$ , the statement $f(x) = g(x)$ for $x = 0$ is FALSE.

Statement 3: $f(x) = g(x)$ for $x = 1$

To determine if this statement is true or false, we need to evaluate the two functions at $x = 1$ .

f(1) = 2^1 + 1 = 3

g(1) = \left(\frac{1}{3}\right)^1 - 2 = \frac{1}{3} - 2 = -\frac{5}{3}

Since $f(1) \neq g(1)$ , the statement $f(x) = g(x)$ for $x = 1$ is FALSE.

Statement 4: $f(x) = g(x)$ for $x = -1$

To determine if this statement is true or false, we need to evaluate the two functions at $x = -1$ .

f(-1) = 2^{-1} + 1 = \frac{1}{2} + 1 = \frac{3}{2}

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

Since $f(-1) \neq g(-1)$ , the statement $f(x) = g(x)$ for $x = -1$ is FALSE.

Problem 2: Inequalities and Absolute Values

Given the functions $f(x) = 2^x + 1$ and $g(x) = \left(\frac{1}{3}\right)^x - 2$ , we need to evaluate the following statements and drag our TRUE or FALSE response to the appropriate box.

Statement 1: $f(x) > g(x)$ for all values of $x$

To determine if this statement is true or false, we need to compare the two functions. We can start by setting the two functions greater than each other:

2^x + 1 > \left(\frac{1}{3}\right)^x - 2

However, this inequality is not true for all values of $x$ . In fact, the two functions are not greater than each other for any value of $x$ . This is because the base of the exponential function on the left-hand side is 2, while the base of the exponential function on the right-hand side is $\frac{1}{3}$ . These two bases are not equal, so the two functions cannot be greater than each other for all values of $x$ .

Therefore, the statement $f(x) > g(x)$ for all values of $x$ is FALSE.

Statement 2: $f(x) > g(x)$ for $x = 0$

To determine if this statement is true or false, we need to evaluate the two functions at $x = 0$ .

f(0) = 2^0 + 1 = 2

g(0) = \left(\frac{1}{3}\right)^0 - 2 = 1 - 2 = -1

Since $f(0) > g(0)$ , the statement $f(x) > g(x)$ for $x = 0$ is TRUE.

Statement 3: $f(x) > g(x)$ for $x = 1$

To determine if this statement is true or false, we need to evaluate the two functions at $x = 1$ .

f(1) = 2^1 + 1 = 3

g(1) = \left(\frac{1}{3}\right)^1 - 2 = \frac{1}{3} - 2 = -\frac{5}{3}

Since $f(1) > g(1)$ , the statement $f(x) > g(x)$ for $x = 1$ is TRUE.

Statement 4: $f(x) > g(x)$ for $x = -1$

To determine if this statement is true or false, we need to evaluate the two functions at $x = -1$ .

f(-1) = 2^{-1} + 1 = \frac{1}{2} + 1 = \frac{3}{2}

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

Since $f(-1) > g(-1)$ , the statement $f(x) > g(x)$ for $x = -1$ is TRUE.

Problem 3: Absolute Values and Inequalities

Given the functions $f(x) = 2^x + 1$ and $g(x) = \left(\frac{1}{3}\right)^x - 2$ , we need to evaluate the following statements and drag our TRUE or FALSE response to the appropriate box.

Statement 1: $|f(x)| = |g(x)|$ for all values of $x$

To determine if this statement is true or false, we need to compare the absolute values of the two functions. We can start by setting the absolute values equal to each other:

|2^x + 1| = |\left(\frac{1}{3}\right)^x - 2|

However, this equation is not true for all values of $x$ . In fact, the absolute values of the two functions are not equal for any value of $x$ . This is because the base of the exponential function on the left-hand side is 2, while the base of the exponential function on the right-hand side is $\frac{1}{3}$ . These two bases are not equal, so the absolute values of the two functions cannot be equal for all values of $x$ .

Therefore, the statement $|f(x)| = |g(x)|$ for all values of $x$ is FALSE.

Statement 2: $|f(x)| = |g(x)|$ for $x = 0$

To determine if this statement is true or false, we need to evaluate the absolute values of the two functions at $x = 0$ .

|f(0)| = |2^0 + 1| = |2| = 2

|g(0)| = |\left(\frac{1}{3}\right)^0 - 2| = |1 - 2| = |-1| = 1

Since $|f(0)| \neq |g(0)|$ , the statement $|f(x)| = |g(x)|$ for $x = 0$ is FALSE.

Statement 3: $|f(x)| = |g(x)|$ for $x = 1$

To determine if this statement is true or false, we need to evaluate the absolute values of the two functions at $x = 1$ .

|f(1)| = |2^1 + 1| = |3| = 3

|g(1)| = |\left(\frac{1}{3}\right)^1 - 2| = |-\frac{5}{3}| = \frac{5}{3}

Since $|f(1)| \neq |g(1)|$ , the statement $|f(x)| = |g(x)|$ for $x = 1$ is FALSE.

Statement 4: $|f(x)| = |g(x)|$ for $x = -1$

To determine if this statement is true or false, we need to evaluate the absolute values of the two functions at $x = -1$ .

|f(-1)| = |2^{-1} + 1| = |\frac{1}{2<br/> **Submit Test: 24-25 Pre-College Math**

Q&A: Functions and Graphs

Q: What is the difference between the two functions $f(x) = 2^x + 1$ and $g(x) = \left(\frac{1}{3}\right)^x - 2$ ?

A: The two functions are different because they have different bases. The base of the exponential function on the left-hand side is 2, while the base of the exponential function on the right-hand side is $\frac{1}{3}$ . This means that the two functions will behave differently as $x$ changes.

Q: How do the two functions compare for $x = 0$ ?

A: For $x = 0$ , we have:

Problem 1: Functions and Graphs

Statement 1: f(x)=g(x)f(x) = g(x)f(x)=g(x) for all values of xxx

Statement 2: f(x)=g(x)f(x) = g(x)f(x)=g(x) for x=0x = 0x=0

Statement 3: f(x)=g(x)f(x) = g(x)f(x)=g(x) for x=1x = 1x=1

Statement 4: f(x)=g(x)f(x) = g(x)f(x)=g(x) for x=−1x = -1x=−1

Problem 2: Inequalities and Absolute Values

Statement 1: f(x)>g(x)f(x) > g(x)f(x)>g(x) for all values of xxx

Statement 2: f(x)>g(x)f(x) > g(x)f(x)>g(x) for x=0x = 0x=0

Statement 3: f(x)>g(x)f(x) > g(x)f(x)>g(x) for x=1x = 1x=1

Statement 4: f(x)>g(x)f(x) > g(x)f(x)>g(x) for x=−1x = -1x=−1

Problem 3: Absolute Values and Inequalities

Statement 1: ∣f(x)∣=∣g(x)∣|f(x)| = |g(x)|∣f(x)∣=∣g(x)∣ for all values of xxx

Statement 2: ∣f(x)∣=∣g(x)∣|f(x)| = |g(x)|∣f(x)∣=∣g(x)∣ for x=0x = 0x=0

Statement 3: ∣f(x)∣=∣g(x)∣|f(x)| = |g(x)|∣f(x)∣=∣g(x)∣ for x=1x = 1x=1

Statement 4: ∣f(x)∣=∣g(x)∣|f(x)| = |g(x)|∣f(x)∣=∣g(x)∣ for x=−1x = -1x=−1

Q&A: Functions and Graphs

Statement 1: $f(x) = g(x)$ for all values of $x$

Statement 2: $f(x) = g(x)$ for $x = 0$

Statement 3: $f(x) = g(x)$ for $x = 1$

Statement 4: $f(x) = g(x)$ for $x = -1$

Statement 1: $f(x) > g(x)$ for all values of $x$

Statement 2: $f(x) > g(x)$ for $x = 0$

Statement 3: $f(x) > g(x)$ for $x = 1$

Statement 4: $f(x) > g(x)$ for $x = -1$

Statement 1: $|f(x)| = |g(x)|$ for all values of $x$

Statement 2: $|f(x)| = |g(x)|$ for $x = 0$

Statement 3: $|f(x)| = |g(x)|$ for $x = 1$

Statement 4: $|f(x)| = |g(x)|$ for $x = -1$