Pre-Calculus pls explain how to do!!?
Find the values of b such that the function has the given maximum value.
f(x) = −x^2 + bx − 12; Maximum value: 109
b=___(smaller value)
b=___(larger value)
Find the values of b such that the function has the given maximum value.
f(x) = −x^2 + bx − 12; Maximum value: 109
b=___(smaller value)
b=___(larger value)
Comments
f(x) = -x² + bx − 12
f′(x) = -2x + b
f(x) will be at maximum when f′(x) = 0
-2x + b = 0
∴ b = 2x
and will have the same values for x and b at -x² + bx − 12 = 109
∴ -x² + (2x)x − 12 = 109
∴ -x² + 2x² − 12 = 109
∴ x² − 12 = 109
∴ x² = 121
∴ x = ±11
when x = -11 then b = 2*-11 = -22 and when x = 11 then b = 22
f(x) = -x² + 22x − 12 then f(11) = 109
and
f(x) = -x² − 22x − 12 then f(-11) = 109
so smaller b = -22
larger b = 22
f(x) = -x^2 + bx - 12
Note that the vertex form of a parabola is y = a(x - h)^2 + k, therefore from completing the square:
f(x) = -(x^2 - bx + 12)
f(x) = -((x - b/2)^2 + 12 - (b^2)/4)
k = (b^2)/4 - 12 = 109
b^2/4 = 121
b^2 = 121*4
b = +/- sqrt(121*4)
= +/- 11*2
= -22 (smaller value)
= +22 (larger value)