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anonymous

  • one year ago

If ƒ(x ) = x 2 + 1 and g(x ) = 3x + 1, find 2 · ƒ(4).

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  1. anonymous
    • one year ago
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    @Nnesha

  2. SolomonZelman
    • one year ago
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    \(\large\color{purple}{ \displaystyle f(x)=x^2+1 }\) \(\large\color{brown}{ \displaystyle g(x)=3x+1 }\) To find \(2\cdot f(2)\) you don't need the \(g(x)\). Just, plug in 2 instead of x into the \(f(x)\), and then multiply the result times two.

  3. Nnesha
    • one year ago
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    4*

  4. SolomonZelman
    • one year ago
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    Ok, if I wanted to find: \(4\cdot f(3)\) then this is what I would do: \(\large\color{black}{ \displaystyle f(x)=x^2+1 }\) Plug in 3 instead of x: \(\large\color{black}{ \displaystyle f(\color{red}{3})=\color{red}{3}^2+1 }\) I know that \(3^2=9\), so I can write that: \(\large\color{black}{ \displaystyle f(3)=9+1 }\) And then, obviously, 9+1=10, so this is what I get for \(f(3)\). \(\large\color{black}{ \displaystyle f(3)=10 }\) Now, I have to multiply: I am asked to find \(4 \cdot f(3)\), and since I know that \(f(3)=10\), I can just go ahead and apply that (substitute 10 for f(3) ): \(\large\color{black}{ \displaystyle 4\cdot f(3)=4 \cdot 10 = 40 }\)

  5. SolomonZelman
    • one year ago
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    (hope this is a helpful example)

  6. anonymous
    • one year ago
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    i got an answer that wasnt one of the options

  7. Nnesha
    • one year ago
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    what did you get ?

  8. anonymous
    • one year ago
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    5.3

  9. SolomonZelman
    • one year ago
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    That is not possible

  10. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle f(x)=x^2+1 }\) \(\large\color{black}{ \displaystyle f(4)=\bf ? }\)

  11. anonymous
    • one year ago
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    i said i did it wrong then i redid it and got it right

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