mathcalculus
Help (Attached below) Determine the extrema of below on the given interval



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mathcalculus
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mathcalculus
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i know we have to find the derivative and set it to zero

mathcalculus
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i got x= 8 and x= 2/9

mathcalculus
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now to have the intervals, im a little lost on...

mathcalculus
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would there only be 2 intervals?

mathcalculus
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or 3? 02/9 then 2/98 and 8infinity?

campbell_st
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ok... so whats the derivative...?

mathcalculus
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9x^274x+16

mathcalculus
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then i set it to = 0

campbell_st
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ok... so what are the solutions...?

mathcalculus
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x=8 and x=2/9

mathcalculus
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critical numbers

campbell_st
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yep... so on [0, 4] substitute x = 0, 2/9 and 4 into the original equation and find the corresponding values...

mathcalculus
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why 4?

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thanks

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isn't 4 a y?

mathcalculus
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and what 8?

mathcalculus
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because the 2 critical numbers are 8 and 2/9

campbell_st
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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

mathcalculus
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ooooooh, 8 would be the other one

mathcalculus
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let me try it

campbell_st
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so the value x =0 is f(0) = 2
x = 2/9 f(2/9) = 3.76
x = 4 f(4) = ?

mathcalculus
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334

mathcalculus
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so the max is 3.76 and min is 334?

campbell_st
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thats correct...

campbell_st
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so just test x = 9, x = 8 and x = 9... and see what happens... I think x = 2/9 remains the max

campbell_st
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its an add interval since x = 9 gives a very large negative number