#### Answer

(a) The speed is a maximum at x = 6.
(b) The particle has a turning point at x = 2 and at x = 8.
(c) The particle could remain at rest at x = 3 and x = 6

#### Work Step by Step

(a) We know that $E = KE + PE$
KE is a maximum when PE is a minimum. PE is a minimum at x = 6, so KE is a maximum at x = 6. Therefore the speed is a maximum at x = 6.
(b) The particle has a turning point at x = 2 and at x = 8.
When PE = E, the kinetic energy KE is zero. The particle turns back to the left at x = 8 and the particle turns back to the right at x = 2.
(c) If E = PE at a point where PE is at a local minimum (the bottom of a potential well), then the particle will stay at rest. If E is changed so that E = PE at x = 3, the particle could remain at rest at x = 3. If E is changed so that E = PE at x = 6, the particle could remain at rest at x = 6.
The particle could remain at rest at x = 3 and x = 6.