How do you find the maximum or minimum of #f(x)=3-x^2-6x#?

2 Answers
AJ
Mar 24, 2018

#(-3,12)#

Explanation:

Since this is a parabolic equation and the value of #a<0# it open downwards so it have a absolute maxima. The maximum point is determined by #x_(max) = -b/(2a)# where b and a are coefficients. This formula is developed using differential calculus.
So,
#x_(max) = -b/(2a)#
#=-(-6)/(2×(-1))#
#=6/(-2)#
#:. X_(max) = -3#.
Now evaluate #f (-3)# get the maximum point.

sankarankalyanam
Mar 24, 2018

It's Maximum

Explanation:

The second derivative demonstrates whether a point with zero first derivative is a maximum, minimum or an inflexion point.

#f(x) = 3 - x^2 - 6x#

#f'(x) = (d/(dx)) (3 - 6x - x^2 = -6 - 2x #

#-6 - 2x = 0 " or" x = -3#

f"(x) #= (d/(dx)) (-2x - 6 ) = -2#

Since f"(x) is negative, it is maximum.

graph{(-x^2 - 6x + 3) [-11.21, 8.79, -13.32, -3.32]}

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