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The curve with equation y= x^5 + 20x^2 - 8 passes through the point P, where x = -2....

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jasonofinn22 | eNotes Newbie

Posted February 23, 2012 at 11:05 PM via web

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The curve with equation y= x^5 + 20x^2 - 8 passes through the point P, where x = -2. Find the value of (d^2)y / dx^2 at the point P.

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted February 23, 2012 at 11:15 PM (Answer #1)

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You need to diferentiate the function with respect to x such that:

`f'(x) = (dy)/(dx) = 5x^4 + 40x`

You need to differentiate the function `(dy)/(dx) = 5x^4 + 40x`  with respect to x such that:

`f"(x) = (d^2y)/(dx^2) = 20x^3 + 40`

You need to substitute -2 for x in equation `(d^2y)/(dx^2) = 20x^3 + 40`  such that:

`f"(-2) = 20*(-2)^3 + 40`

`f"(-2) = - 160 + 40 =gt f"(-2) = -120`

Hence, evaluating `(d^2y)/(dx^2)`  at `x = -` 2 yields  `f"(-2) = -120.`

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cheezea | Student, Grade 10 | (Level 1) Valedictorian

Posted March 1, 2012 at 9:13 PM (Answer #2)

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dy/dx = 5x^4 + 40x

(d^2)y / dx^2 = 20x^3 +40

Sub x= -2,

(d^2)y / dx^2

= 20(-2)^3 +40

= -120 :)

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