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

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Answer 1 · 2017-01-22T03:02:58

At #(-3, 12)#the function has a maximum.

Given -

#y=3-x^2-6x#

Take the first derivative

#dy/dx=-2x-6#

Find for what value of #x; dy/dx# becomes zero

#dy/dx=0 => -2x-6=0#

#-2x-6=0#
#-2x=6#
#x=6/(-2)=-3#

Its interpretation is when #x# takes the value #0#; the slope of the curve is zero. The curve turns.

To find the point of turn, substitute #x=-3# in the given function

At #x=-3#

#y=3-(-3)^2-6(-3)#
#y=3-9+18=12#

At #(-3, 12)#the curve's slope is zero. The curve truns.

It may be minimum point or maximum point.

To find this calculate the second derivative

#(d^2y)/(dx^2)=-2 <0#

The second derivative is less than zero. Hence the curve has a maximum.

At #(-3, 12)#the function has a maximum.

graph{3-x^2-6x [-32.48, 32.48, -16.24, 16.24]}