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Question 2 of 9 , Step 1 of 1 2//15 Correct Find (f@g)(1) for the following functions. Round your answer to two decimal places, if necessary. f(x)=1+sqrtx" and "g(x)=sqrt(x^(2)+24) Answer How to enter your answer (opens in new window) (f@g)(1)=

\newlineFind (f∘g)(1) (f \circ g)(1) for the following functions. Round your answer to two decimal places, if necessary.\newlinef(x)=1+x and g(x)=x2+24 f(x)=1+\sqrt{x} \text { and } g(x)=\sqrt{x^{2}+24} \newlineAnswer\newlineHow to enter your answer (opens in new window)\newline(f∘g)(1)= (f \circ g)(1)=

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Q. \newlineFind (f∘g)(1) (f \circ g)(1) for the following functions. Round your answer to two decimal places, if necessary.\newlinef(x)=1+x and g(x)=x2+24 f(x)=1+\sqrt{x} \text { and } g(x)=\sqrt{x^{2}+24} \newlineAnswer\newlineHow to enter your answer (opens in new window)\newline(f∘g)(1)= (f \circ g)(1)=
  1. Find g(1)g(1): To find (f∘g)(1)(f\circ g)(1), we first need to evaluate g(1)g(1) and then plug that result into the function ff.
  2. Evaluate g(1)g(1): Evaluate g(1)g(1) by substituting xx with 11 in the function g(x)=x2+24g(x) = \sqrt{x^2 + 24}.
    g(1)=12+24g(1) = \sqrt{1^2 + 24}
    g(1)=1+24g(1) = \sqrt{1 + 24}
    g(1)=25g(1) = \sqrt{25}
    g(1)=5g(1) = 5
  3. Substitute into ff: Now that we have g(1)=5g(1) = 5, we substitute this value into the function ff to find (f@g)(1)(f@g)(1).\newline(f@g)(1)=f(g(1))=f(5)(f@g)(1) = f(g(1)) = f(5)
  4. Evaluate f(5)f(5): Evaluate f(5)f(5) by substituting xx with 55 in the function f(x)=1+xf(x) = 1 + \sqrt{x}.\newlinef(5)=1+5f(5) = 1 + \sqrt{5}\newlinef(5)=1+2.236f(5) = 1 + 2.236 (rounded to three decimal places for intermediate calculation)\newlinef(5)=3.236f(5) = 3.236 (rounded to three decimal places for intermediate calculation)
  5. Round final answer: Round the final answer to two decimal places as instructed.\newline(f@g)(1)=3.24(f@g)(1) = 3.24 (rounded to two decimal places)

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