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#1

$Old$ January 2nd, 2009, 03:59 PM

steveo0 $steveo0 is offline$

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$Default$ FTC w/ Chain Rule and Concavity

The graph of f(x) = [0,x] integral (15t^2 - 2t^3 + 24) dt is concave up on (a,b). Find b - a

so would f(x) = 15x^2 - 1/2x^4 + 24x?
and then I take the derivative of that which would be 15x^2 - 2x^3 + 24 and set it equal to zero? this is confusing . . .

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$Old$ January 2nd, 2009, 04:08 PM

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$f(x) = \int_0^x 15t^2 - 2t^3 + 24 \, dt$

$f'(x) = 15x^2 - 2x^3 + 24$

$f''(x) = 30x - 6x^2$

if f(x) is concave up on (a,b), then f''(x) > 0 on (a,b)

$30x - 6x^2 > 0$

$6x(5 - x) > 0$

f''(x) > 0 on (0,5) ... $b-a = 5$

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#3

$Old$ January 2nd, 2009, 04:42 PM

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oh shoot! i forgot second derivative was to determine concavity. man winter break haha. thanks !

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