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# 4.1.14-eg - f x is undeﬁned it suﬃces to look at the...

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Example Find all critical points of the function f ( x ) = x + 1 x . Solution: Recall that the critical points of a function f ( x ) are those values of x in the domain of f for which f 0 ( x ) = 0 or f 0 ( x ) is undefined. Since f ( x ) = x 1 / 2 + x - 1 / 2 , it follows that f 0 ( x ) = 1 2 x - 1 / 2 - 1 2 x - 3 / 2 = 1 2 x 1 / 2 - 1 2 x 3 / 2 . Finding a common denominator, we have f 0 ( x ) = x - 1 2 x 3 / 2 . (1) To determine the values of x for which f 0 ( x ) = 0, we can look at the values of x which make the numerator in (1) zero, i.e., x - 1 = 0 or x = 1 . To determine the values of x
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Unformatted text preview: f ( x ) is undeﬁned, it suﬃces to look at the values of x which make the denominator in (1) zero, or for which 2 x 3 / 2 = 0 , or x = 0 . Note however that even though f is undeﬁned at x = 0, the original function f is not deﬁned at that point either, so x = 0 is not a critical point for f . So the function f has only one critical point, at x = 1 ....
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