QuestionSuppose that the functions $g$ and $h$ are defined for all real numbers $x$ as follows. $\begin{array}{l} g(x)=2 x+1 \\ h(x)=3 x \end{array}$
Write the expressions for $(g+h)(x)$ and $(g \cdot h)(x)$ and evaluate $(g-h)(1)$ . $\begin{array}{r} (g+h)(x)=\square \\ (g \cdot h)(x)=\square \\ (g-h)(1)=\square \end{array}$

Studdy Solution

STEP 1

1. The functions $g(x) = 2x + 1$ and $h(x) = 3x$ are defined for all real numbers $x$ .
2. We are asked to find expressions for the sum $(g+h)(x)$ , the product $(g \cdot h)(x)$ , and evaluate the difference $(g-h)(1)$ .

STEP 2

1. Find the expression for $(g+h)(x)$ .
2. Find the expression for $(g \cdot h)(x)$ .
3. Evaluate $(g-h)(1)$ .

STEP 3

To find $(g+h)(x)$ , we add the functions $g(x)$ and $h(x)$ :
$(g+h)(x) = g(x) + h(x)$
Substitute the given functions:
$(g+h)(x) = (2x + 1) + (3x)$
Combine like terms:
$(g+h)(x) = 2x + 3x + 1 = 5x + 1$

STEP 4

To find $(g \cdot h)(x)$ , we multiply the functions $g(x)$ and $h(x)$ :
$(g \cdot h)(x) = g(x) \cdot h(x)$
Substitute the given functions:
$(g \cdot h)(x) = (2x + 1) \cdot (3x)$
Distribute $3x$ through the terms in $(2x + 1)$ :
$(g \cdot h)(x) = 3x \cdot 2x + 3x \cdot 1$
$(g \cdot h)(x) = 6x^2 + 3x$

STEP 5

To evaluate $(g-h)(1)$ , we first find $(g-h)(x)$ :
$(g-h)(x) = g(x) - h(x)$
Substitute the given functions:
$(g-h)(x) = (2x + 1) - (3x)$
Combine like terms:
$(g-h)(x) = 2x + 1 - 3x = -x + 1$
Now evaluate at $x = 1$ :
$(g-h)(1) = -1 + 1 = 0$
The expressions and evaluation are: $\begin{array}{r} (g+h)(x) = 5x + 1 \\ (g \cdot h)(x) = 6x^2 + 3x \\ (g-h)(1) = 0 \end{array}$

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