Question 8

Given the function $f(x) = \frac{2x + 1}{3x - 5}$ .

Based on the information above, determine whether each of the following statements is TRUE or FALSE.

$f^{-1}(x) = \frac{5x + 1}{3x - 2}$ .
The value $y = \frac{5}{3}$ is not a member of the range of $y = f^{-1}(x)$ .
The value $x = \frac{1}{2}$ satisfies $f(x) = f^{-1}(x)$ .

The correct sequence of truth values for statements 1, 2, and 3 is ....

Let's check the truth of each statement.

We will find the inverse function of $f(x) = \frac{2x + 1}{3x - 5}$ .

Let $f(x) = y$ , then:

y = \frac{2x + 1}{3x - 5}

y(3x - 5) = 2x + 1

3xy - 5y = 2x + 1

3xy - 2x = 5y + 1

x(3y - 2) = 5y + 1

x = \frac{5y + 1}{3y - 2}

So, the inverse function is:

f^{-1}(x) = \frac{5x + 1}{3x - 2}

Statement 1 is True.

The range of the inverse function $f^{-1}(x)$ is equal to the domain of the original function $f(x)$ .

The function $f(x) = \frac{2x + 1}{3x - 5}$ is defined if the denominator is not equal to zero:

3x - 5 \neq 0

3x \neq 5

x \neq \frac{5}{3}

This means the domain of $f(x)$ is all real numbers except $\frac{5}{3}$ .

Therefore, the range of $f^{-1}(x)$ is all real numbers except $\frac{5}{3}$ .

So, the value $y = \frac{5}{3}$ is indeed not a member of the range of $f^{-1}(x)$ .

Statement 2 is True.

We check if $x = \frac{1}{2}$ satisfies $f(x) = f^{-1}(x)$ .

Calculate $f\left(\frac{1}{2}\right)$ :

f\left(\frac{1}{2}\right) = \frac{2(\frac{1}{2}) + 1}{3(\frac{1}{2}) - 5}

= \frac{1 + 1}{\frac{3}{2} - \frac{10}{2}}

= \frac{2}{-\frac{7}{2}}

= 2 \times \left(-\frac{2}{7}\right) = -\frac{4}{7}

Calculate $f^{-1}\left(\frac{1}{2}\right)$ :

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

= \frac{\frac{5}{2} + \frac{2}{2}}{\frac{3}{2} - \frac{4}{2}}

= \frac{\frac{7}{2}}{-\frac{1}{2}}

= \frac{7}{2} \times (-2) = -7

Since $-\frac{4}{7} \neq -7$ , then $x = \frac{1}{2}$ does not satisfy the equation.

Statement 3 is False.

The correct sequence of truth values is True, True, False.

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