7) find two functions $f$ and $g$ such that $h(x)=(f\\circ g)(x)$. \n$h(x)=\\sqrt5{8x-3}$\n○ a) $f(x)=x-3$…

7) find two functions $f$ and $g$ such that $h(x)=(f\\circ g)(x)$. \n$h(x)=\\sqrt5{8x-3}$\n○ a) $f(x)=x-3$ and $g(x)=\\sqrt5{8x}$\n○ b) $f(x)=\\sqrt5{x}$ and $g(x)=8x-3$\n○ c) $f(x)=8x-3$ and $g(x)=\\sqrt5{x}$\n○ d) $f(x)=\\sqrt5{8x}$ and $g(x)=x-3$

7) find two functions $f$ and $g$ such that $h(x)=(f\\circ g)(x)$. \n$h(x)=\\sqrt5{8x-3}$\n○ a) $f(x)=x-3$ and $g(x)=\\sqrt5{8x}$\n○ b) $f(x)=\\sqrt5{x}$ and $g(x)=8x-3$\n○ c) $f(x)=8x-3$ and $g(x)=\\sqrt5{x}$\n○ d) $f(x)=\\sqrt5{8x}$ and $g(x)=x-3$

Answer

Explanation:

Step1: Recall function composition rule

$(f\circ g)(x) = f(g(x))$

Step2: Test Option A

Substitute $g(x)$ into $f(x)$: $f(g(x)) = \sqrt[5]{8x} - 3 \neq \sqrt[5]{8x-3}$

Step3: Test Option B

Substitute $g(x)$ into $f(x)$: $f(g(x)) = \sqrt[5]{8x - 3} = h(x)$

Step4: Verify other options (optional)

Option C: $f(g(x)) = 8\sqrt[5]{x} - 3 \neq \sqrt[5]{8x-3}$ Option D: $f(g(x)) = \sqrt[5]{8(x-3)} = \sqrt[5]{8x-24} \neq \sqrt[5]{8x-3}$

Answer:

B) $f(x) = \sqrt[5]{x}$ and $g(x) = 8x - 3$