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November 28, 2014

November 28, 2014

Posted by **Marissa** on Tuesday, April 19, 2011 at 10:58pm.

f(-1)=3 and f'(-1)=6

let h(x)=x^5f(x)

evaluate h'(x) at x=-1

- Math- Calculus -
**Mgraph**, Tuesday, April 19, 2011 at 11:08pmh'(x)=5*x^4*f(x)+x^5*f'(x)

h'(-1)=5*(-1)^4*f(-1)+(-1)^5*f'(-1)=

5*1*3+(-1)*6=15-6=9

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