How do you find the maximum or minimum of #f(x)=-7-3x^2+12x#?

1 Answer
Narad T.
Oct 30, 2016

The maximum is at #(2,5)#

Explanation:

We calculate the derivative
#f(x)=-7-3x^2+12x#
The domain of #f(x)# is #RR#
#f'(x)=-6x+12#
For a maximum or minimum, #f'(x)=0#
so, #-6x+12=0# #=>##6x=12#
#x=2#
To see if it's a max or min, we make a chart
#x##color(white)(aaaa)##-oo##color(white)(aaaa)##2##color(white)(aaaa)##+oo#
#f'(x)##color(white)(aaaa)##+##color(white)(aaaa)##-#
#f(x)##color(white)(aaaaa)##uarr##color(white)(aaaa)##darr#

So we have a maximum at #(2,5)#
we can also look at the sign of the second derivative
#f''(x)=-6# #(< 0)# this is a max
graph{-3x^2+12x-7 [-12.66, 12.65, -6.33, 6.33]}

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