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The derivative of f is given f'(x) = e^x(-x^3+3x)-3 for [0,5].

+1 vote

What value of f(x) is an absolute minimum?

asked Mar 30, 2013 in CALCULUS by andrew Scholar

1 Answer

+2 votes

The  derivative of f is given f1(x) = ex(-x3 + 3x) - 3 for interval [0 , 5]

f1(x) = e1(-x3 + 3x) - 3

Absolute minimum value  of f(x)  : put x = o in  f1(x)

f1(0) = e0(-03 + 3(0)) - 3

f1(0) = 1(0 + 0) - 3

f1(0) = 0 - 3

f1(0) = -3

There fore absolute minimum value of f(x) : -3.

 

 

 

     

 

 

answered Mar 30, 2013 by diane Scholar

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