3. Suppose the derivative of g(t) is g′(t) = (t − 3) (t − 2) (t − 8). g(t) has a local maximum at t = 3 because Possibilities: (a) g(t) 0 to the immediate left of t = 3 and g(t) 0 to the immediate left of t = 3 and g′(t)

Question

3. Suppose the derivative of g(t) is g′(t) = (t − 3) (t − 2) (t − 8). g(t) has a local maximum at t = 3 because Possibilities: (a) g(t) > 0 to the immediate left of t = 3 and g(t) < 0 to the immediate right of t = 3 (b) g′(3) = 0 (c) g′(t) < 0 to the immediate left of t = 3 and g′(t) > 0 to the immediate right of t = 3 (d) g(t) < 0 to the immediate left of t = 3 and g(t) > 0 to the immediate right of t = 3 (e) g′(t) > 0 to the immediate left of t = 3 and g′(t) < 0 to the immediate right of t = 3

anonymous · Answer

the answer is E. why?

anonymous · Answer

Well remember that in order to find the critical points you first have to set the derivative (g'(t)) equal to zero. Now, you take that point and plug it into the second derivative and if it is negative, then it is a max, if it is positive, then it is a min.

anonymous · Answer

can you use a visual or something? i get the gist of it but not 100 good on these kinds of problems.

anonymous · Answer

(t − 3) (t − 2) (t − 8)=0 you should find that t=2,3,8
Take the second derivative...
$$3t ^{2}-26t+46$$
Plug in 3 and and you will get 3(9)-(26*3)+46=-5 
It is a negative, so it is a max.