Applied Calculus

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Let f(x) = −x2 + ax + b.
Determine the constants a and b such that f has a relative maximum at x = 2 and the relative maximum value is 13.

  • Applied Calculus -

    f'(x) = -2x+a
    if f'=0 at x=2, then a=4

    f(x) = -x^2 + 4x + b
    f(2) = 13 = -4+8+b
    b=5

    f(x) = -x^2 + 4x + 5 = -(x+1)(x-5)

    vertex is at x=(5-1)/2 = 2 as desired

  • Applied Calculus -

    Correction: b=9

    f(x)=-x^2 +4x+b

    if f(2)=13=-2^2+4(2)+b
    then f(2)=13=-4+8+b
    17=8+b
    9=b

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