Question

suppose that the functions \\(s\\) and \\(t\\) are defined for all real numbers \\(x\\) as follows.

\\s(x) = 3x + 1\\
\\t(x) = 4x\\

find the following.

\\(s - t)(x) = \\
\\(s \cdot t)(x) = \\
\\(s + t)(1) = \\

Explanation:

Define the given functions

We are given two functions defined for all real numbers \(x\):

$$s(x) = 3x + 1$$

$$t(x) = 4x$$

Find the difference function \((s - t)(x)\)

Using the Algebraic Expressions concept, we subtract the function \(t(x)\) from \(s(x)\):

$$(s - t)(x) = s(x) - t(x)$$

$$(s - t)(x) = (3x + 1) - (4x)$$

$$(s - t)(x) = -x + 1$$

Find the product function \((s \cdot t)(x)\)

Using the Algebraic Expressions concept, we multiply the two functions:

$$(s \cdot t)(x) = s(x) \cdot t(x)$$

$$(s \cdot t)(x) = (3x + 1)(4x)$$

$$(s \cdot t)(x) = 12x^2 + 4x$$

Find the sum evaluated at 1, \((s + t)(1)\)

First, we find the sum function \((s + t)(x)\):

$$(s + t)(x) = s(x) + t(x) = (3x + 1) + 4x = 7x + 1$$

Now, we evaluate this sum function at \(x = 1\):

$$(s + t)(1) = 7(1) + 1 = 8$$

Answer:

Suppose that the functions \(s\) and \(t\) are defined for all real numbers \(x\) as follows.

$$s(x) = 3x + 1$$

$$t(x) = 4x$$

Find the following.
\((s - t)(x) =\) <blank>\(-x + 1\)</blank>
\((s \cdot t)(x) =\) <blank>\(12x^2 + 4x\)</blank>
\((s + t)(1) =\) <blank>8</blank>