3. Suppose the derivative of H(s) is given by H\'(s) = (s^2 + 3)(s^2 + 7). Find

Question

3. Suppose the derivative of H(s) is given by H'(s) = (s^2 + 3)(s^2 + 7). Find the value of s in the interval [-10, 10] where H(s) takes on its maximum. Possibilities: (a)7 (b)3 (e) -7 (d) -10 (e) 10 4. Suppose the derivative of g(t) is g'(t) = -9(t - 4)(t - 8). For t in which interval(s) is g concave up? Possibilities: (a) (-infinity,6) (b) (6,infinity) (c) (-9, 4) U (6, 8) (d) (-infinity,4) U (8,infinity) (e) (4,8) 5. Suppose the derivative of h(x) is given by h'(x) = (x - 3) (x - 7). If h(x) is concave upward on the interval (a, infinity), what is a? Possibilities: (a)3 (b)7 (c) -infinity (d)5 (e) 10

Explanation / Answer

H'(s) = (s^2 + 3)(s^2 + 7)

Now, we need to find critical values

(s^2 + 3)(s^2 + 7) = 0

s^2 = -3 , s^2 = -7

So, no solution

And therefore, no critical numbers exist

Which means, we need to just plug in the endpoints of the given interval to find the max

H'(s) = s^4 + 10s^2 + 21

Integrating :

H(s) = (1/5)s^5 + (10/3)s^3 + 21s + C

Now, since we do not have any conditon, it means that we cannot definitely state the value of C

Now, the endpoints of the interval are -10 and 10

The leading term (1/5)*s^5 becomes very positive when s = 10 and and very negative when s = -10

So, with a good degree of sureness, we can say that the max is attained at s = 10

Option E

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g'(t) = -9(t - 4)(t - 8)

Multiplying it out :

g'(t) = -9t^2 + 108t - 288 = 0

For concavity, we need to derive again

g''(t) = -18t + 108

First we need the inflection points ----> -18t + 108 =0 ----> t = 6

This splits the domain into (-inf , 6) and (6 , inf)

Region 1 : (-inf , 6)
Testvalue = 0
g''(t) = -18t + 108
g''(0) = 0 + 108 = 108 --> positive
So, CONCAVE UP

Region 2 : (6 , inf)
Testvalue = 7
g''(7) = -18(7) + 108 --> -18 --> negative
So, CONCAVE DOWN on this region

So, answer for concave up : (-inf , 6) ----> ANSWER

Option A

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h'(x) = (x - 3)(x - 7) = x^2 - 10x + 21

Deriving again :

h''(x) = 2x - 10 = 0

2x = 10

x = 5 ---> this is the inflection point

This splits the domain into (-inf , 5) and (5 , inf)

Already given that the concave up interval is (a , inf)

Comparing (5 , inf) to (a , inf), we get :

a = 5 ---> ANSWER

Option D

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