The function $$ f(x)=\frac{5 x+6}{x+5} $$ is one-to-one. (a) Find its inverse and check your answer. (b) Find the domain and the range of $$ f $$ and $$ f^{-1} $$. (a) $$ f^{-1}(x)=\square $$ (Simplify your answer.)

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**Step 1. Set up the equation**

We start with the function

$$
f(x)=\frac{5x+6}{x+5}.
$$

Let $$ y = f(x) $$, so

$$
y = \frac{5x+6}{x+5}.
$$

To find the inverse function, we swap $$ x $$ and $$ y $$:

$$
x = \frac{5y+6}{y+5}.
$$

---

**Step 2. Solve for $$ y $$**

Multiply both sides by $$ y+5 $$ to eliminate the denominator:

$$
x(y+5) = 5y+6.
$$

Expand the left side:

$$
xy + 5x = 5y + 6.
$$

Bring the terms involving $$ y $$ to one side:

$$
xy - 5y = 6 - 5x.
$$

Factor $$ y $$ on the left side:

$$
y(x-5) = 6 - 5x.
$$

Solve for $$ y $$ by dividing both sides by $$ x-5 $$ (note that $$ x\neq 5 $$):

$$
y = \frac{6-5x}{x-5}.
$$

Thus, the inverse function is

$$
f^{-1}(x)=\frac{6-5x}{x-5}.
$$

---

**Step 3. Check the Inverse**

To check, we compute $$ f(f^{-1}(x)) $$:

Start with

$$
f\Bigl(\frac{6-5x}{x-5}\Bigr)=\frac{5\left(\frac{6-5x}{x-5}\right)+6}{\left(\frac{6-5x}{x-5}\right)+5}.
$$

Simplify the numerator:

$$
5\left(\frac{6-5x}{x-5}\right)+6 = \frac{5(6-5x)}{x-5} + \frac{6(x-5)}{x-5} 
= \frac{30-25x+6x-30}{x-5} 
= \frac{-19x}{x-5}.
$$

Simplify the denominator:

$$
\left(\frac{6-5x}{x-5}\right)+5 = \frac{6-5x}{x-5} + \frac{5(x-5)}{x-5} 
= \frac{6-5x+5x-25}{x-5} 
= \frac{-19}{x-5}.
$$

Now, the whole expression becomes

$$
\frac{\frac{-19x}{x-5}}{\frac{-19}{x-5}} = x.
$$

Since we obtain $$ f(f^{-1}(x))=x $$, the inverse function is verified.

---

**Step 4. Domain and Range**

- **For $$ f(x)=\frac{5x+6}{x+5} $$:**

- **Domain of $$ f $$:**
  
    The denominator cannot be zero, so
  
    $$
    x+5 \neq 0 \quad \Longrightarrow \quad x\neq -5.
    $$
  
    Therefore, the domain of $$ f $$ is
  
    $$
    \{x\in \mathbb{R} \mid x\neq -5\}.
    $$
  
  - **Range of $$ f $$:**
  
    As $$ x\to \pm\infty $$, the function approaches the horizontal asymptote
    
    $$
    y=5.
    $$
    
    To check if $$ y=5 $$ is achieved, set:
  
    $$
    \frac{5x+6}{x+5}=5.
    $$
    
    Multiply both sides by $$ x+5 $$:
    
    $$
    5x+6 = 5x+25.
    $$
    
    This simplifies to:
    
    $$
    6=25,
    $$
    
    which is impossible. Thus, $$ y=5 $$ is not included in the range.
    
    Therefore, the range of $$ f $$ is
  
    $$
    \{y\in \mathbb{R} \mid y\neq 5\}.
    $$

- **For the inverse $$ f^{-1}(x)=\frac{6-5x}{x-5} $$:**

- **Domain of $$ f^{-1} $$:**
  
    The denominator requires that
  
    $$
    x-5 \neq 0 \quad \Longrightarrow \quad x\neq 5.
    $$
  
    Thus, the domain of $$ f^{-1} $$ is
  
    $$
    \{x\in \mathbb{R} \mid x\neq 5\}.
    $$
  
    Note that this is exactly the range of $$ f $$.
  
  - **Range of $$ f^{-1} $$:**
  
    The range of $$ f^{-1} $$ is the domain of $$ f $$, so:
  
    $$
    \{y\in \mathbb{R} \mid y\neq -5\}.
    $$

---

**Final Answers:**

(a) The inverse function is

$$
f^{-1}(x)=\frac{6-5x}{x-5}.
$$

(b) The domain and range are:

- For $$ f(x)=\frac{5x+6}{x+5} $$:
  - Domain: $$\{x\in \mathbb{R} \mid x\neq -5\}$$
  - Range: $$\{y\in \mathbb{R} \mid y\neq 5\}$$

- For $$ f^{-1}(x)=\frac{6-5x}{x-5} $$:
  - Domain: $$\{x\in \mathbb{R} \mid x\neq 5\}$$
  - Range: $$\{y\in \mathbb{R} \mid y\neq -5\}$$
(a) The inverse function is $$ f^{-1}(x) = \frac{6 - 5x}{x - 5} $$. (b) - **Domain and Range of $$ f(x) = \frac{5x + 6}{x + 5} $$:** - Domain: All real numbers except $$ x = -5 $$. - Range: All real numbers except $$ y = 5 $$. - **Domain and Range of $$ f^{-1}(x) = \frac{6 - 5x}{x - 5} $$:** - Domain: All real numbers except $$ x = 5 $$. - Range: All real numbers except $$ y = -5 $$.

The function \( f(x)=\frac{5 x+6}{x+5} \) is one-to-one. (a) Find its inverse and check your answer. (b) Find the domain and the range of \( f \) and \( f^{-1} \). (a) \( f^{-1}(x)=\square \) (Simplify your answer.)

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