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### 5.2 first and second derivative test

• 1. A Summary of Curve Sketching
• 2. What should you consider? • • • • • • • • • symmetries x-intercept y-intercept relative extrema asymptores concavity inflection points intervals of increase interval of decrease ASYMPTOTE – VA HA SA N(x) ax m Let R(x) = = D(x) bx n VERTICAL ASYMPTOTE If D(x) = 0, then VA : x = c HORIZONTAL ASYMPTOTE If m > n, then HA : doesn't exist If m < n, then HA : y = 0, If m = n, then HA : y = SLANT ASYMPTOTE If x - axis a b m = n +1, then SA : y = mx + b Long divide N(x) by D(x)
• 3. Let’s see how that works!!! 2x 2 - 8 Sketch the graph of the equation f (x) = 2 x -16 2x - 8 =0 2 x -16 2 VA: 2x 2 - 8 = 0 x 2 -16 = 0 f '(x) = f '(x) = 2 × 02 - 8 1 f (0) = 2 = 0 -16 2 x = -2, 2 x = -4, 4 HA: 4x ( x 2 -16) - 2x ( 2x 2 - 8) (x (x 2 -16) 16 ( x - 4) 2 -16) ( x 2 -16) - f '(x) = 2 = 16 ( x - 4) · -4 · 4 4x 3 - 64 - 4x 3 +16x (x 2 -16) f '(x) = ( x + 4) ( x - 4) ( x 2 -16) - y=2 m=n + 2 = 16x - 64 (x 2 -16) 2 16 =0 2 ( x + 4) ( x -16)
• 4. Almost Done !!!! f "(x) = -16 é( x 2 -16) + 2x ( x + 4)ù ë û é( x + 4) ( x -16)ù ë û 2 2 -16 é x -16 + 2x + 8xù ë û 2 f "(x) = f "(x) = =0 2 é( x + 4) ( x 2 -16)ù ë û 2 -16 (3x - 4) ( x + 4) ( x + 4 ) ( x + 4) ( x - 2 -16) · -4 2 =0 =0 f "(x) = -16 é3x 2 + 8x -16ù ë û é( x + 4) ( x -16)ù ë û f "(x) = 2 2 -16 (3x - 4) ( x + 4) ( x - · + · 4/3 4 =0 2 -16) 2 =0
• 5. Let’ make a list • • • • • • • • • -2, 2 x-intercept 0.5 y-intercept -4, 4 VA 2 HA ________ SA [ 4, +¥) f(x) ( -¥, 4] f(x) ______ X max X min 4 • f(x) È • f(x) Ç • X infl æ4 ö ç , 4÷ è3 ø æ 4ö -¥, ÷ and ( 4, +¥) ç è 3ø 4 3
• 6. Sketch the graph of the equation • • • • • • • • • x-intercept y-intercept VA HA SA f(x) f(x) X max X min • f(x) • f(x) • X infl y = ( x - 4) 2 3
• 7. 1 3 Sketch the graph of the equation y = 6x + 3x • • • • • • • • • x-intercept y-intercept VA HA SA f(x) f(x) X max X min • f(x) • f(x) • X infl 4 3
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