Question

1. let $f(x) = x + 3$, $g(x) = 4x^2$, and $h(x) = \sqrt{x - 7}$. find the indicated value. $h(g(2))$ 7 let $f(x) = 2x - 5$, $g(x) = x^{-2}$, and $h(x) = 3x + 4$. perform the indicated composition and state the domain. $h(f(x))$ domain:

Explanation:

Problem 4: Find \( h(g(2)) \)

Step 1: Find \( g(2) \)

We know that \( g(x) = 4x^2 \). Substitute \( x = 2 \) into \( g(x) \):
\( g(2)=4\times(2)^2 = 4\times4 = 16 \)

Step 2: Find \( h(g(2)) \)

Now, we need to find \( h(16) \) since \( g(2) = 16 \). We know that \( h(x)=\sqrt{x - 7} \). Substitute \( x = 16 \) into \( h(x) \):
\( h(16)=\sqrt{16 - 7}=\sqrt{9} = 3 \)

Step 1: Find \( h(f(x)) \)

We know that \( f(x)=2x - 5 \) and \( h(x)=3x + 4 \). To find \( h(f(x)) \), we substitute \( f(x) \) into \( h(x) \) wherever we see \( x \) in \( h(x) \):
\( h(f(x))=h(2x - 5)=3(2x - 5)+4 \)
Simplify the expression:
\( h(f(x)) = 6x-15 + 4=6x-11 \)

Step 2: Find the domain of \( h(f(x)) \)

First, we consider the domain of the inner function \( f(x) \). The function \( f(x)=2x - 5 \) is a linear function, and the domain of a linear function is all real numbers, i.e., \( (-\infty,\infty) \).

Then, we consider the domain of the outer function \( h(x) \). The function \( h(x)=3x + 4 \) is also a linear function, and its domain is all real numbers, i.e., \( (-\infty,\infty) \).

Since there are no restrictions from the composition (the output of \( f(x) \) is a real number for all real \( x \), and \( h(x) \) is defined for all real numbers), the domain of \( h(f(x)) \) is all real numbers.

Answer:

\( 3 \)

Problem 7: Find \( h(f(x)) \) and its domain