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

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

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

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

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

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

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

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

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

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

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

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
thanks
 11 months ago

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

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

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

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
let me try it
 11 months 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) = ?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
334
 11 months ago

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

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