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SECOND DERIVATIVE TEST

SECOND DERIVATIVE TEST. The graph of f ( x ) will have a minimum at x = a , if f ′ ( a ) = 0, and f ″ ( x ) > 0 The graph o f f ( x) will have a maximum at x = a , if f ′ ( a ) = 0, and f ″ ( x ) < 0

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SECOND DERIVATIVE TEST

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  1. SECOND DERIVATIVE TEST • The graph of f(x) will have a minimum at x = a, if f ′ (a) = 0, and f ″ (x) > 0 • The graph off(x) will have a maximum at x = a, if f ′ (a) = 0, and f ″ (x) < 0 • The graph off(x) may have an inflection point at x = a, if f ′ (a) = 0, and f ″(x) = 0

  2. EXAMPLE 1g(x) = 2x3 – 54x– 48 Step 1: Find first derivative g′ (x) = 6x2 – 54 Step 2: Set first derivative equal to 0 and solve for x 6x2 – 54 = 0 6x2 = 54 x2 = 9 Critical values or turning points are at x = ± 3

  3. EXAMPLE 1g(x) = 2x3 – 48x – 48 Step 3: Find the second derivative g″ (x) = 12x Step 4: Substitute values for x in Step 2 into the second derivative (3, – 156) is a minimum point Since Local minimum value is – 156 (–3, 60) is a maximum point Since Local maximum value is 60

  4. EXAMPLE 1g(x) = 2x3 – 48x – 54 (-3 , 60) (3 , -156)

  5. x = 0 or x = ¾ critical values EXAMPLE 2 f(x) = x4 – x3 f”(x) = 12x2 – 6x f ′(x) = 4x3 – 3x2 f "(0) = 12(0)2 – 6(0) = 0 4x3 – 3x2= 0 x2 (4x – 3)= 0 f(0) = 0 is an inflection point and f " ( ¾ ) = 12( ¾ ) 2 – 6( ¾ ) = 9/4 f( ¾ ) = -0.11 is a local minimum

  6. EXAMPLE 2 f(x) = x4 – x3 (0 , 0) (3/4 , 9/4)

  7. x = 0, x = + 1 critical values EXAMPLE 3 h(x) = 3x5 – 5x3 f ′ (x) = 15x4 – 15x2 h" (x) = 60x3 – 30x 15x4 – 15x2 = 0 15x2(x2 – 1) = 0 h"(0) = 60(0)3 – 30(0) = 0 15x2(x – 1)(x + 1) = 0 (0, 0) is an inflection point h"(1) = 60(1)3 – 30(1) = 30 (1, –2) is a local minimum h"(–1) = 60(–1)3– 30(–1) = –30 (–1, 2) is a local maximum

  8. EXAMPLE 3 h(x) = 3x5 – 5x3 (1 , 2) (0 , 0) (1 , -2)

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