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7 September, 06:52

If f (x) = 2x^2+1 and g (x) = x^2-7 find (f+g) (x)

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Answers (2)
  1. 7 September, 06:54
    0
    F (x) + g (x) = [2x^2 + 1] + [x^2 - 7]

    ≈ 3x^2 - 6

    Hence (f + g) (x) = 3x^2 - 6
  2. 7 September, 07:14
    0
    Answer:f=-

    7



    x

    2

    + 1-2g-x



    Step-by-step explanation:

    1 Subtract {x}^{2}x

    2

    from both sides.

    2{x}^{2}+1-g-x-{x}^{2}=-7f+g2x

    2

    + 1-g-x-x

    2

    = -7f+g

    2 Simplify 2{x}^{2}+1-g-x-{x}^{2}2x

    2

    + 1-g-x-x

    2

    to {x}^{2}+1-g-xx

    2

    + 1-g-x.

    {x}^{2}+1-g-x=-7f+gx

    2

    + 1-g-x=-7f+g

    3 Subtract gg from both sides.

    {x}^{2}+1-g-x-g=-7fx

    2

    + 1-g-x-g=-7f

    4 Simplify {x}^{2}+1-g-x-gx

    2

    + 1-g-x-g to {x}^{2}+1-2g-xx

    2

    + 1-2g-x.

    {x}^{2}+1-2g-x=-7fx

    2

    + 1-2g-x=-7f

    5 Divide both sides by - 7-7.

    -/frac{{x}^{2}+1-2g-x}{7}=f-

    7



    x

    2

    + 1-2g-x

    = f

    6 Switch sides.

    f=-/frac{{x}^{2}+1-2g-x}{7}f=-

    7



    x

    2

    + 1-2g-x
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