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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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