# 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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