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**Maximum ??? Minimum??? How can we tell?**

If f is increasing just to the left of a critical number c and decreasing 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 and increasing just to the right of c, then f has a local minimum at c

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First Derivative Test Let c be a critical number of a continuous function f. If f ' (x) changes from positive to negative at c, then f has a local maximum at c.

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First 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.

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First 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.

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**EXAMPLE 1: f(x) = 3x2 – 4x + 13 f ′(x) = 6x – 4 6x – 4 < 0**

𝑥< 2 3 𝑥> 2 3 𝑥= 2 3 (critical number) f ′(x) < 0 f ′(x) > 0 tangent slope is negative tangent slope is positive 𝒇 𝟐 𝟑 = 𝟑𝟓 𝟑 =𝟏𝟏.𝟕 local minimum at

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**EXAMPLE 2: f(x) = x3 – 12x – 5 f ′(x) = 3x2 – 12 3x2 – 12 = 0**

Test for x < –2 3( – 2)( ) 3(x2 – 4) = 0 Test for –2 < x < 2 3( – 2)( ) 3(x – 2)(x + 2) = 0 x = 2 or x = –2 Critical values Test for x > 2 3( – 2)( ) f (–2) 11 max min f (2) = -21 + + – 2 2 Local maximum at value is 11 Local minimum value is -21

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**EXAMPLE 3 f(x) = x4 – x3 f ′(x) = 4x3 – 3x2 f (0) = 0 f ( ¾ ) = 0.11**

Test for x < 0 ( )2 (4( ) – 3) 4x3 – 3x2 = 0 x2 (4x – 3) = 0 Test for 0 < x < ¾ ( )2 (4( ) – 3) critical values x = 0 or x = ¾ Test for x > ¾ ( )2 (4( ) – 3) + Local maximum DNE Local minimum value is 0.11

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DIVIDING INTEGERS 1. IF THE SIGNS ARE THE SAME THE ANSWER IS POSITIVE 2. IF THE SIGNS ARE DIFFERENT THE ANSWER IS NEGATIVE.

DIVIDING INTEGERS 1. IF THE SIGNS ARE THE SAME THE ANSWER IS POSITIVE 2. IF THE SIGNS ARE DIFFERENT THE ANSWER IS NEGATIVE.

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