step 2 collect all terms involving dy dx respect to

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Unformatted text preview: critical points and the intervals of increase/decrease. Step 4. Find inflection points and the intervals in which the function is concave up/down. EXAMPLE: Given a function with the property ekx (2kx 2 − 1) , x2 2 f ′( x) = k k > 0 , determine the intervals on which f is increasing and decreasing and the location of any local maximum and minimum values of f. SOLUTION: We find that e kx (2kx 2 − 1) = 0 when x = ±1 / 2k , x2 2 f '( x) = k hence these points are the locations of any local maximum or minimum values of f. Now we examine the change of the sign of the derivative of f at these points. has a relative minimum at p. () If f ′( p ) = 0 and f ′′ p < 0 , then f has a relative maximum at p. If f ′( p ) = 0 and f ′′( p ) = 0 , then the test fails. f ′′( p ) > 0 on an interval, then f(x) is concave up on that interval; if f ′′( p ) < 0 , then Concavity: If x −1/ More free study sheet ′′and practice tests=at: 2k Differentiate both sides: 2x + y + x ⋅ 2 Where is the curve...
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This note was uploaded on 03/15/2010 for the course MAT MAT135 taught by Professor Treung during the Fall '08 term at University of Toronto.

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