Questions: Suppose that the functions (s) and (t) are defined for all real numbers (x) as follows [ s(x)=x-2 t(x)=4 x+3 ] Write the expressions for ((s cdot t)(x)) and ((s-t)(x)) and evaluate ((s+t)(2)). [ (s cdot t)(x)=square (s-t)(x)=square (s+t)(2)=square ]

Suppose that the functions (s) and (t) are defined for all real numbers (x) as follows
[
s(x)=x-2 
t(x)=4 x+3
]

Write the expressions for ((s cdot t)(x)) and ((s-t)(x)) and evaluate ((s+t)(2)).
[
(s cdot t)(x)=square 
(s-t)(x)=square 
(s+t)(2)=square
]
Transcript text: Suppose that the functions $s$ and $t$ are defined for all real numbers $x$ as foll \[ \begin{array}{l} s(x)=x-2 \\ t(x)=4 x+3 \end{array} \] Write the expressions for $(s \cdot t)(x)$ and $(s-t)(x)$ and evaluate $(s+t)(2)$. \[ \begin{array}{r} (s \cdot t)(x)=\square \\ (s-t)(x)=\square \\ (s+t)(2)=\square \end{array} \]
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Solution

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Solution Steps

Step 1: Determine \((s \cdot t)(x)\)

To find \((s \cdot t)(x)\), we need to multiply the functions \(s(x)\) and \(t(x)\):

\[ s(x) = x - 2 \] \[ t(x) = 4x + 3 \]

The product \((s \cdot t)(x)\) is given by:

\[ (s \cdot t)(x) = s(x) \cdot t(x) = (x - 2)(4x + 3) \]

Expanding this expression:

\[ = x(4x + 3) - 2(4x + 3) \] \[ = 4x^2 + 3x - 8x - 6 \] \[ = 4x^2 - 5x - 6 \]

Step 2: Determine \((s-t)(x)\)

To find \((s-t)(x)\), we need to subtract the function \(t(x)\) from \(s(x)\):

\[ (s-t)(x) = s(x) - t(x) = (x - 2) - (4x + 3) \]

Simplifying this expression:

\[ = x - 2 - 4x - 3 \] \[ = x - 4x - 2 - 3 \] \[ = -3x - 5 \]

Step 3: Evaluate \((s+t)(2)\)

To find \((s+t)(2)\), we need to add the functions \(s(x)\) and \(t(x)\) and then evaluate at \(x = 2\):

\[ s(x) = x - 2 \] \[ t(x) = 4x + 3 \]

The sum \((s+t)(x)\) is:

\[ (s+t)(x) = s(x) + t(x) = (x - 2) + (4x + 3) \]

Simplifying this expression:

\[ = x - 2 + 4x + 3 \] \[ = 5x + 1 \]

Now, evaluate at \(x = 2\):

\[ (s+t)(2) = 5(2) + 1 = 10 + 1 = 11 \]

Final Answer

\[ \begin{array}{r} (s \cdot t)(x) = \boxed{4x^2 - 5x - 6} \\ (s-t)(x) = \boxed{-3x - 5} \\ (s+t)(2) = \boxed{11} \end{array} \]

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