150 f(3)=0 f'(2)=32 ∫(4)=-32 100 f(1)=64 16x-+96x f(5)=-64 f(x)=-32x+96 50 f"(0=96 f(6)=-96 FIGURE 3. 8 Copyright@2011 Pearson Education, Inc. Publishing as Pears on Addison-Wesley Slide 3-14Copyright © 2011 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 3 - 14