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Graphing Using Derivatives

Graphing Using Derivatives. Step 1. find the 1 st derivative set it equal to zero solve it. f ( x ) = x 3 - 3 x 2. f '( x ) = 3 x 2 - 6 x = 3 x ( x -2) = 0 x = 0, x = 2 . Step 2. Making the 1 st derivative chart:

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Graphing Using Derivatives

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  1. Graphing Using Derivatives

  2. Step 1 • find the 1st derivative • set it equal to zero • solve it f(x) = x3 - 3x2 f '(x) = 3x2 - 6x = 3x (x -2) = 0 x = 0, x = 2

  3. Step 2 • Making the 1st derivative chart: • Plug the 1st derivative x values into the 1st derivative equation f '(x) = 3x2 - 6x = 3x (x - 2) X=0, x=2 Pick a value to the left of x=0 such as –2 3x2 - 6x = 3(-2)2 – 6(-2) = 24 (positive number)

  4. If the first derivative f' is positive (+) , then the function f is increasing ( ). 2. If the first derivative f' is negative (-) , then the function f is decreasing ( ). • 1st derivative determines where the graph is increasing/decreasing and critical points • -The graph is decreasing for 0 < x < 2 • The graph is increasing for x < 0 and x > 2

  5. Step 3-Identify relative max, relative min, & terrace points Relative maxima Relative minima + 0 - = relative max - 0 + = relative min A terrace point occurs when both signs are the same: + 0 + or - 0 -

  6. Step 4 • Find the ordered pair of the critical points To find the ordered pair of the critical points, ALWAYS PLUG INTO THE ORIGINAL EQUATION f(x) = x3 - 3x2 Relative min: f(x) = x3 - 3x2 = 03 - 3(0)2 = 0 (0,0) Relative min: f(x) = x3 - 3x2 = 23 - 3(2)2 = - 4 (2,-4)

  7. Step 5 • find the 2nd derivative • set it equal to zero • solve it f '(x) = 3x2 - 6x f ''(x) = 6x – 6 = 6(x - 1) = 0 x = 1

  8. Step 6 • Making the 2nd derivative chart: • Plug the 2nd derivative x values into the 2nd derivative equation This time, the (-) and (+) do not determine increasing/decreasing, but where the graph is concave up and concave down 2nd derivative is used to find inflection points, or where 2nd derivative=0 and there is a change in concavity

  9. Step 7-Identify Inflection Points f is concave up for x > 1 ; f is concave down for x < 1 To find the inflection point, you take x = 1 and plug it into the ORIGINAL EQUATION f(x) = x3 - 3x2 = 13 - 3(1)2 = -2 (1,-2)

  10. More practice 1st derivative: F(x) = 4x/x2 +1 x = 1 x = - 1

  11. …More Practice 2nd derivative: = 0 x = 0 Concave up Concave down Concave down Concave up

  12. Information found from each chart: From f ’ : • f is increasing for –1 < x < 1 • f is decreasing for x < -1 and x > 1 • From f ’ : • f is concave up for • f is concave down for • f has inflection points at: x = 0 y = 0

  13. Congratulations! Now you’re ready to make a graph!

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