First Derivative Test for Identifying Local Maximum and Minimum
This mathematical concept explains how to find local maximum and minimum points of a continuous function using the first derivative test. If the function is decreasing just to the right of a critical
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Slide1Maximum ??? Minimum???How can we tell? and decreasing just to the right of c , then f has a local minimum at c If f is increasing just to the left of a critical number c and increasing just to the right of c , then f has a local maximum at c If f is decreasing just to the left of a critical number c
Slide2First Derivative Test Let c be a critical number of a continuous function f . 1. If f ' ( x ) changes from positive to negative at c , then f has a local maximum at c .
Slide3First Derivative Test Let c be a critical number of a continuous function f . 2. If f ' ( x ) changes from negative to positive at c , then f has a local minimum at c.
Slide4First Derivative Test Let c be a critical number of a continuous function f . 3. If f ' ( x ) does not change sign at c , then f has no maximum or minimum at c.
Slide5EXAMPLE 1: f ( x ) = 3 x 2 – 4 x + 13 f ′( x ) = 6 x – 4 6 x – 4 = 0 ( critical number) f ′( x ) < 0 local minimum at 6 x – 4 < 0 6 x < 4 6 x – 4 > 0 6 x > 4 tangent slope is positive tangent slope is negative f ′( x ) > 0
Slide6EXAMPLE 2: f ( x ) = x 3 – 12 x – 5 f ′( x ) = 3 x 2 – 12 3 x 2 – 12 = 0 3( x 2 – 4) = 0 3( x – 2)( x + 2) = 0 x = 2 or x = –2 – 2 2 + + Test for x < –2 3( – 2)( + 2) Test for –2 < x < 2 3( – 2)( + 2) Test for x > 2 3( – 2)( + 2) max min Local minimum value is -21 f (–2) 11 f (2) = -21 Local maximum at value is 11 Critical values
Slide7EXAMPLE 3 f ( x ) = x 4 – x 3 f ′( x ) = 4 x 3 – 3 x 2 4 x 3 – 3 x 2 = 0 x 2 (4 x – 3) = 0 critical values x = 0 or x = ¾ f (0) = 0 f ( ¾ ) = 0.11 Test for x < 0 ( ) 2 (4( ) – 3) Test for 0 < x < ¾ ( ) 2 (4( ) – 3) Test for x > ¾ ( ) 2 (4( ) – 3) 0 + Local minimum value is 0.11 Local maximum DNE