Second+Derivative+Test+-+Wed,+Mar.+28


 * Today’s Agenda:**
 * 1) Got test back
 * 2) Took up homework question (txb. – p. 331 # 13)
 * 3) Did a similar, but more plausible test question
 * 4) Lesson: Second Derivative Test
 * 5) Announcement: Quiz Friday


 * txb. – p. 331 # 13 (txb had the wrong solution/question)**


 * A similar, but more plausible test question**



Have you ever wanted to know for sure whether an x-value you found from the first derivative was really a local maximum or minimum…well…this lesson is the SOLUTION for you!
 * Second Derivative Test Lesson Introduction:**

**__ Second Derivative Test __**

Given the function f(x), where f’(c) = 0 and f’’(x) exists on the interval containing c.

If f’’(c) > 0 then f(c) is a local minimum. If f’’(c) < 0 then f(c) is a local maximum. If f’’(c) = 0 then use the first derivative test.



Find the local maximum and minimums.
 * Example:**

f(x) = 2x3 – 24x

//Find the first derivative// f’(x) = 6x2-24

//Set it to zero and find the value(s) for x// 0 = 6(x2-4) 0 = 6x(x-2)(x+2) x = - 2, x = 2

//Find the second derivative// f’’(x) = 12x

//Sub in the x-values from the first derivative to see whether it is a local maximum/minimum.// f’’(-2) = -24 concave down (maximum) f’’(2) = 24 concave up (minimum)

//Plug the x-values from the first derivative back into the original equation to find the y-values.//

Therefore, (-2, 32) is a local maximum and (2, -32) is a local minimum.


 * If you are still having trouble…these are some extra links you may check out:**

1. Video (more of an intro to lesson)

http://www.youtube.com/watch?v=wRBCvDy2jEY

2. Web page (similar to what we did in class):

http://revisionworld.co.uk/a2-level-level-revision/maths/pure-mathematics/calculus/second-derivative


 * Practice at…**

txb – p. 332 #15