Problem 16 - Algebra and Trigonometry

Question

Evaluate the function at the indicated values. $$ \begin{array}{l}{h(t)=t+\frac{1}{t}} \\ {h(1), h(-1), h(2), h\left(\frac{1}{2}\right), h(x), h\left(\frac{1}{x}\right)}\end{array} $$

Step-by-Step Solution

Answer

h(1) = 2, h(-1) = -2, h(2) = 2.5, h(1/2) = 2.5, h(x) = x + 1/x, h(1/x) = x + 1/x.

Step 1: Evaluate at t = 1

Substitute \( t = 1 \) into the function: \[ h(1) = 1 + \frac{1}{1} \]Simplify the expression: \[ h(1) = 1 + 1 = 2 \] Thus, \( h(1) = 2 \).

Step 2: Evaluate at t = -1

Substitute \( t = -1 \) into the function:\[ h(-1) = -1 + \frac{1}{-1} \]Simplify the expression:\[ h(-1) = -1 - 1 = -2 \]Thus, \( h(-1) = -2 \).

Step 3: Evaluate at t = 2

Substitute \( t = 2 \) into the function:\[ h(2) = 2 + \frac{1}{2} \]Simplify the expression:\[ h(2) = 2 + 0.5 = 2.5 \]Thus, \( h(2) = 2.5 \).

Step 4: Evaluate at t = 1/2

Substitute \( t = \frac{1}{2} \) into the function:\[ h\left(\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{\frac{1}{2}} \]Simplify the expression:\[ h\left(\frac{1}{2}\right) = \frac{1}{2} + 2 \]\[ h\left(\frac{1}{2}\right) = 0.5 + 2 = 2.5 \]Thus, \( h\left(\frac{1}{2}\right) = 2.5 \).

Step 5: Evaluate at t = x

Substitute \( t = x \) into the function:\[ h(x) = x + \frac{1}{x} \]This is the general formula for \( h(t) \) so no simplification is needed. Thus, \( h(x) = x + \frac{1}{x} \).

Step 6: Evaluate at t = 1/x

Substitute \( t = \frac{1}{x} \) into the function:\[ h\left(\frac{1}{x}\right) = \frac{1}{x} + \frac{1}{\left(\frac{1}{x}\right)} \]Simplify the expression:\[ h\left(\frac{1}{x}\right) = \frac{1}{x} + x \]Thus, \( h\left(\frac{1}{x}\right) = x + \frac{1}{x} \).

Key Concepts

Algebra and Function Evaluation

Algebra involves manipulating mathematical symbols and solving equations to find unknown values. When it comes to function evaluation, this involves substituting a given input value into a function and calculating the output. Functions act like rules; for each input value, there is an output that's directly determined by the function's formula. In our exercise, the function given is: \[ h(t) = t + \frac{1}{t} \]This formula shows a simple rational expression, where a variable is both in the denominator and the numerator. To evaluate this function for specific input values, the variable \( t \) is replaced with the value in question. Each substitution may yield a different result, depending on the inherent operations in the function.

The Art of Substitution

Substitution is a key technique in algebra, allowing us to replace a variable with a specific value or another expression. This is fundamental when evaluating functions, as it enables us to compute the output for different inputs. Let's break it down:

• Identify the variable in the function you need to substitute. In our exercise, that's \( t \).
• Replace every instance of this variable in the function with the given value.
• Ensure the replacement is precise throughout the entire expression.

For example, when we substitute \( t = 1 \) into the function \( h(t) = t + \frac{1}{t} \), the expression becomes \( h(1) = 1 + \frac{1}{1} \). Here, we substitute each instance of \( t \) with 1. This method of substitution can be applied to evaluate the function at any desired value.

Simplifying Expressions

Once substitution is performed, the next step is simplification. Simplifying expressions involves performing any arithmetic operations and reducing the expression to its simplest form. The goal is to make the expression as straightforward and compact as possible. Here are some pointers on how to simplify:

• Perform basic arithmetic operations like addition, subtraction, multiplication, and division.
• Look for opportunities to combine like terms or diminish fractions to their simplest form.
• Address any complex fractions by simplifying them to ordinary numbers.

In our exercise, after substituting \( t = -1 \) into the function, we get \( h(-1) = -1 + \frac{1}{-1} \). The expression simplifies to \( h(-1) = -1 - 1 = -2 \). Each substitution and simplification reinforces the algebra principles and shows how different values affect the output of a function.