Question

read each question carefully and answer all questions

1. let $f(x) = 3x^2 + 6$ and $g(x) = x + 5$.

part a. find $f(g(x))$.
$f(g(x)) = \square$
part b. state the domain.
a. $\bigcirc$ all real numbers
b. $\bigcirc$ all real numbers except $x = 6$
c. $\bigcirc$ all real numbers except $x = 5$
d. $\bigcirc$ $x \geq 0$

Explanation:

Part A

Step1: Substitute \( g(x) \) into \( f(x) \)

We know \( f(x) = 3x^2 + 6 \) and \( g(x) = x + 5 \). To find \( f(g(x)) \), we replace \( x \) in \( f(x) \) with \( g(x) = x + 5 \). So we get \( f(g(x)) = 3(x + 5)^2 + 6 \).

Step2: Expand \( (x + 5)^2 \)

Using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \), where \( a = x \) and \( b = 5 \), we have \( (x + 5)^2 = x^2 + 10x + 25 \). Then \( f(g(x)) = 3(x^2 + 10x + 25) + 6 \).

Step3: Distribute the 3

Multiply each term inside the parentheses by 3: \( 3x^2 + 30x + 75 + 6 \).

Step4: Combine like terms

Add 75 and 6: \( 3x^2 + 30x + 81 \).

The function \( f(g(x)) = 3x^2 + 30x + 81 \) is a quadratic function. Quadratic functions are polynomials, and the domain of all polynomial functions (including quadratics) is all real numbers because there are no restrictions (like division by zero or square roots of negative numbers) that would exclude any real number \( x \).

Answer:

\( 3x^2 + 30x + 81 \)

Part B