f(x)=2x^(3)+8 h(x)=root(3)(12-5x) Write (f@h)(x) as an expression in terms of x. (f@h)(x)=

Substitute h(x) into f(x): To find the composition of the functions (f@h)(x), we need to substitute the function h(x) into the function f(x). This means we will replace every instance of x in f(x) with h(x).Write down h(x): First, let's write down the function h(x): h(x) = root(3)(12-5x). This is the cube root of (12-5x).Write down f(x): Now, let's write down the function f(x): f(x) = 2x^(3) + 8.Cube root of (12-5x): Substitute h(x) into f(x) to get (f@h)(x): (f@h)(x) = f(h(x)) = 2(h(x))^(3) + 8.Replace cubed h(x): Now we need to cube the expression for h(x): (root(3)(12-5x))^(3). When we cube a cube root, they cancel each other out, so we are left with the expression inside the cube root, which is 12-5x.Simplify the expression: Replace the cubed h(x) in the expression for (f@h)(x): (f@h)(x) = 2(12-5x) + 8.Multiply through the expression: Now we simplify the expression: (f@h)(x) = 2 * 12 - 2 * 5x + 8.Combine like terms: Multiply through the expression: (f@h)(x) = 24 - 10x + 8.Final expression: Combine like terms to get the final expression for (f@h)(x): (f@h)(x) = 32 - 10x.

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f(x)=2x^(3)+8

h(x)=root(3)(12-5x)
Write
(f@h)(x) as an expression in terms of
x.

(f@h)(x)=

$f(x)=2 x^{3}+8$ $\newline$ $h(x)=\sqrt[3]{12-5 x}$ $\newline$ Write $(f \circ h)(x)$ as an expression in terms of $x$ . $\newline$ $(f \circ h)(x)=$

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Math Problems

Algebra 1

Compare linear and exponential growth

Full solution

f(x)=2 x^{3}+8

\newline

h(x)=\sqrt[3]{12-5 x}

\newline

Write

(f \circ h)(x)

as an expression in terms of

x

\newline

(f \circ h)(x)=

Substitute $h(x)$ into $f(x)$ : To find the composition of the functions $(f \circ h)(x)$ , we need to substitute the function $h(x)$ into the function $f(x)$ . This means we will replace every instance of $x$ in $f(x)$ with $h(x)$ .
Write down $h(x)$ : First, let's write down the function $h(x)$ : $h(x) = \sqrt[3]{12-5x}$ . This is the cube root of $(12-5x)$ .
Write down $f(x)$ : Now, let's write down the function $f(x)$ : $f(x) = 2x^{3} + 8$ .
Cube root of $(12-5x)$ : Substitute $h(x)$ into $f(x)$ to get $(f@h)(x)$ : $(f@h)(x) = f(h(x)) = 2(h(x))^{3} + 8$ .
Replace cubed $h(x)$ : Now we need to cube the expression for $h(x)$ : $\left(\sqrt[3]{12-5x}\right)^{3}$ . When we cube a cube root, they cancel each other out, so we are left with the expression inside the cube root, which is $12-5x$ .
Simplify the expression: Replace the cubed $h(x)$ in the expression for $(f@h)(x)$ : $(f@h)(x) = 2(12-5x) + 8$ .
Multiply through the expression: Now we simplify the expression: $(f@h)(x) = 2 \times 12 - 2 \times 5x + 8$ .
Combine like terms: Multiply through the expression: $(f@h)(x) = 24 - 10x + 8$ .
Final expression: Combine like terms to get the final expression for $(f@h)(x)$ : $(f@h)(x) = 32 - 10x$ .