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mathcalculus
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Help (Attached below) Determine the extrema of below on the given interval
 one year ago
 one year ago
mathcalculus Group Title
Help (Attached below) Determine the extrema of below on the given interval
 one year ago
 one year ago

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mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i know we have to find the derivative and set it to zero
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i got x= 8 and x= 2/9
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
now to have the intervals, im a little lost on...
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
would there only be 2 intervals?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
or 3? 02/9 then 2/98 and 8infinity?
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
ok... so whats the derivative...?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
9x^274x+16
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
then i set it to = 0
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
ok... so what are the solutions...?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
x=8 and x=2/9
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
critical numbers
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
yep... so on [0, 4] substitute x = 0, 2/9 and 4 into the original equation and find the corresponding values...
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
why 4?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
thanks
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
isn't 4 a y?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
and what 8?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
because the 2 critical numbers are 8 and 2/9
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
well (a) is asking for the location of that min and max on the interval x = 0 to x = 4 and as x = 2/9 is a critical point...you don't know its nature
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
ooooooh, 8 would be the other one
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
let me try it
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
so the value x =0 is f(0) = 2 x = 2/9 f(2/9) = 3.76 x = 4 f(4) = ?
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
so the max is 3.76 and min is 334?
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
thats correct...
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
so just test x = 9, x = 8 and x = 9... and see what happens... I think x = 2/9 remains the max
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
its an add interval since x = 9 gives a very large negative number
 one year ago
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