Due Wednesday by 11:59pm Points 12 Submitting an external tool HW 2.7 Progress saved Done $ Vo O . .. Score: 0/12 0/12 answered . Question 1 0/1 pt 9 4 2 99 0 Details Suppose that the position of a particle is given by s = f(t) = 5t + 6t + 9. (a) Find the velocity at time t. v(t) = (b) Find the velocity at time t = 3 seconds. m (c) Find the acceleration at time t. a(t) = m (d) Find the acceleration at time t = 3 seconds. m $2 Question Help: Video Add Work Submit QuestionHW 2.7 Progress saved Done VoCO . . . Score: 0.67/12 0/12 answered . Question 10 > 0/1 pt 9 4 2 99 0 Details Given f(x) = 5x3 - 30x2 + 6x + 12 The critical values of f' '(x) are x = Concave Down interval is O Concave Down interval does not exist. O Concave Up interval is O Concave Up interval does not exist. Add Work Submit QuestionHW 2.7 Progress saved Done ... Score: 0.67/12 0/12 answered . Question 11 C 0/1 pt 9 4 2 99 0 Details (+1,4) (2-2) Intervals of x-values where f(x) is concave up Intervals of x-values where f(x) is concave down Add Work Submit QuestionHW 2.7 Progress saved Done Vo ... Score: 0.67/12 0/12 answered . Question 12 0/1 pt 9 4 2 99 0 Details Determine the open intervals on which the graph is concave up or concave down. If there is more than one answer, separate them with a comma. + 4 g(x) = 4 - a) g(x ) is undefined when x = b) g'(x) = *Simplify it. c) g(x) is concave up on and concave down on Add Work Submit QuestionDue Wednesday by 11:59pm Points 12 Submitting an external tool HW 2.7 Progress saved Done $ VO O : Score: 0/12 0/12 answered ts . Question 2 > 0/1 pt 0 4 2 99 0 Details Let s(t) = 6to + 9t - 360t be the equation of motion for a particle. Find a function for the velocity. v(t) = Where does the velocity equal zero? [Hint: factor out the GCF.] t = and t = Find a function for the acceleration of the particle. a(t) = Question Help: @ Video Add Work Submit QuestionDue Wednesday by 11:59pm Points 12 Submitting an external tool HW 2.7 Progress saved Done $ vo O : Score: 0.67/12 0/12 answered . Question 3 > 0.67/1 pt 9 4 # 95 0 Details The function graphed above is decreasing on the interval 0/1 pt 9 4 2 99 0 Details 3 4 5 6 -35 -70 105 140 175 210 245 280 315 350 For the above quartic polynomial f( x ) = - 2x - 15.75x2, identify the interval on which f is concave down. 0/1 pt 9 4 99 0 Details Grades 10- Zoom 8 A 2. -6 -5 -4 -3/-2 1 2 3 4 5 6 12. -10 The function graphed above is: Concave up on the interval(s) Concave down on the interval(s) There is an inflection point at: Question Help: @ Video Add Work Submit QuestionHw 2-7 Progress saved' Done I a '5' 5 Score: 0' 67.112 0112 answered | .Question 6 v | 'TT' a 0/1 pt '0 4 3 99 G) Details Let 39(2) : 23 + 622 , 36:; + a. (a) Use the definition of a derivative or the derivative rules to find 1' 1;...) = :i (b) Use the definition of a derivative or the derivative ruies to find ME) = :1 (c) On what interval is f increasing? (d) On what interval is f decreasing? (e) On what interval is f concave downward? (f) On what interval is f concave upward? ! Add Work 1 Submit Question Hw 2-7 Progress savedi Dane i 6 1m" 3 Score: 0.67.112 0l12 answered | .Question 7 v | 'ZE' 120/1 pt '0 4 3 99 (D Details Let f[a:) : 10z2 + I + 5. (a) Use the definition of a derivative or the derivative rules to find f ' (I) = l:] (b) Use the definition of a derivative or the derivative rules to find 1" '6\") = :1 (c) On what interval is f increasing? (d) On what interval is f decreasing? (e) On what interval is f concave downward? (f) On what interval is f concave upward? | Add Work i Submit Question HW 2-7 Progress saved Done 6 '5' E Score: 0'671'12 0/12 answered 0 Question 8 v B 011 pt '0 4 :3 99 (D Details Let f(;c) = 92:10 41:9. (a) Use the definition of a derivative or the derivative rules to find f M = |:] (b) Use the definition of a derivative or the derivative rules to find f ' '(r) : 1:] (c) On what interval is f increasing? (d) On what interval is f decreasing? (e) On what interval is f concave downward? (f) On what interval is f concave upward? Add Work Submit Question HW 2.7 Progress saved Done & Vo . . . Score: 0.67/12 0/12 answered . Question 9 0/1 pt 0 4 2 99 0 Details 19 -x - 7x'+ 2x+8, C'(2) = 2 19 Given C(x) = x2 - 14x + 2, and C"(x) = 19x - 14 Critical values of C' '(x) are x = Concave Down interval is O Concave Down interval does not exist. O Concave Up interval is O Concave Up interval does not exist. Add Work Submit