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Find the x-values where f ' (x) = 0.

f(x)= (1/3)x^3 - x + 1

asked Mar 19, 2013 in CALCULUS by andrew Scholar

1 Answer

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f(x) = (1/3)x3 - x + 1

Diferenciate each side with respective x  for function : f(x) = (1/3)x3 - x + 1

Recall: derivative of  x^n  = d/dx (xn) = nxn-1  Derivative of constant  = 0

There fore f1(x) = (1/3)3x3-1 - 1x1-1 + 0

Simplify

f1(x) = x2 - x0 + 0

Recall: x0 = 1

f1(x) = x2 - 1

But f1(x) = 0

Substitute f1(x) = 0

x2 - 1 = 0

Add 1 to each side

x2 -1 +1 = 0 +1

x2 + 0 = 1

x2 = 1

Take square root each side

x = -1, 1

x values is -1, 1.

answered Mar 19, 2013 by diane Scholar

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