{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}