Consider the function

h(x)= 2x+1/x+5

If necessary, give answers as fractions in simplest for. For example: -3/2

Find the following.

**(a) h(11) = Answer 1 Question 9** 


**(b) 3+2h(11) = Answer 2 Question 9** 


**(c) State which value of *x* must be excluded from any domain of h(x). Answer 3 Question 9** 

**(d) Solve h(x) = 1**

**x = Answer 4 Question 9** 

Let's break down the problem step by step for each part.

The function given is:

\[
h(x) = \frac{2x + 1}{x + 5}
\]

### **(a) Finding \( h(11) \):**

We need to substitute \( x = 11 \) into the function:

\[
h(11) = \frac{2(11) + 1}{11 + 5} = \frac{22 + 1}{16} = \frac{23}{16}
\]

So, the answer to **(a)** is:

\[
\boxed{\frac{23}{16}}
\]

### **(b) Finding \( 3 + 2h(11) \):**

From part (a), we know that \( h(11) = \frac{23}{16} \). Now, substitute that into the expression:

\[
3 + 2h(11) = 3 + 2 \times \frac{23}{16} = 3 + \frac{46}{16}
\]

Convert 3 to a fraction with a denominator of 16:

\[
3 = \frac{48}{16}
\]

Now add:

\[
3 + 2h(11) = \frac{48}{16} + \frac{46}{16} = \frac{94}{16}
\]

Simplify the fraction:

\[
\frac{94}{16} = \frac{47}{8}
\]

So, the answer to **(b)** is:

\[
\boxed{\frac{47}{8}}
\]

### **(c) Excluding values from the domain of \( h(x) \):**

The function \( h(x) \) is undefined when the denominator is zero. The denominator is \( x + 5 \), so we set it equal to zero to find the excluded value:

\[
x + 5 = 0 \quad \Rightarrow \quad x = -5
\]

So, the value of \( x \) that must be excluded from the domain is:

\[
\boxed{-5}
\]

### **(d) Solving \( h(x) = 1 \):**

We need to solve the equation \( h(x) = 1 \), meaning:

\[
\frac{2x + 1}{x + 5} = 1
\]

Multiply both sides by \( x + 5 \) to eliminate the denominator:

\[
2x + 1 = (x + 5)
\]

Simplify the equation:

\[
2x + 1 = x + 5
\]

Subtract \( x \) from both sides:

\[
x + 1 = 5
\]

Now subtract 1 from both sides:

\[
x = 4
\]

So, the solution to \( h(x) = 1 \) is:

\[
\boxed{4}
\]

### Summary of Answers:

**(a)** \( h(11) = \frac{23}{16} \)

**(b)** \( 3 + 2h(11) = \frac{47}{8} \)

**(c)** The value \( x = -5 \) must be excluded from the domain.

**(d)** The solution to \( h(x) = 1 \) is \( x = 4 \).

Would you like more details on any part? Here are 5 related questions to explore further:

1. How can we determine the range of the function \( h(x) \)?
2. What is the behavior of the function \( h(x) \) as \( x \to -5 \)?
3. Can you find the inverse of the function \( h(x) \)?
4. What is the derivative of the function \( h(x) \)?
5. What are the critical points of the function \( h(x) \)?

**Tip:** Always check for values that make the denominator zero when dealing with rational functions.

Consider the function h(x) = (2x + 1) / (x + 5). If necessary, give answers as fractions in simplest form. For example: -3/2. Find the following. (a) h(11) = Answer 1 Question 9. (b) 3 + 2h(11) = Answer 2 Question 9. (c) State which value of x must be excluded from any domain of h(x). Answer 3 Question 9. (d) Solve h(x) = 1. x = Answer 4 Question 9.

Learn how to solve the rational function h(x) = (2x + 1) / (x + 5) by evaluating h(11), finding 3 + 2h(11), determining domain exclusions, and solving h(x) = 1. This problem covers key algebraic concepts for Grades 9-11.

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