M4i9 13 D Pathways IMathAS Assess x C Chegg Study Guided Sox

M4i9 #13

D Pathways IMathAS Assess x C Chegg Study Guided Sox 4 C A https://imathas.rationalreasoning.net/assessment/showtest.php?action=skip&to;=12 Q & D & O G ® E book rks b rt book Home > Murdock Precalc _013 > Assessment Jeremee Hemphill M4 19 Homework Due Wed 0427 2016 11:59 pm Show Intro Instructions Questions Suppose j is an exponential function. Use the table below to answer the following questions . Q 1 Q2 [44 Q3 [6 6, Q4 [3/3 VQ 5 VQ 6 (3/3 Q7 [6 6, Q 8 4/4 Q9 (3/3 Q 10 [1/1] Q 11 (03) Q 12 (05 Q 13 (03) Q 14 (03) C Q 15 (03) Q 16 03 Q 17 (0/6) Q 18 (0/4) Q 19 (0/1) F 0.5 .6 2. (f) 1.4 0.98 (0.686 (0.33614 a. Evaluate j(0) Preview b. What is the 1-unit growth factor? Preview C. Evaluate j (114 ) Preview Points possible: 3 Unlimited attempts. Post this question to forum a Submit Grade: 32/63 Print version He I bp O A Cl 2:47 PM 4/26/2016

Solution

Solution:

Let j(f) = ab^f

when f = 0.5, then j(f) = ab^0.5 =1.4

and when f = 1, then j(f) = ab^1 =0.98

Dividing both the equations, we get

ab^1/ab^0.5 = 0.98/1.4

b^(0.5) = 0.7

or   b = 0.7^2

           =0.49.

We also have ab^1 =0.98

or      a(0.49) = 0.98

or        a = 2.

Therefore the function is : j(f) = 2 * 0.49^f

Thus j(0) = 2*0.49^0 = 2.                                         Ans.

b. When f increase by 1 unit then j(f) decreases by

0.686-1.4 = -0.714 unit

Therefore growth factor is -0.714 unit .                     Ans.

c. We know the function is : j(f) = 2 * 0.49^f   or   f = j^-(2 * 0.49^f)

Thus 2 * 0.49^f =144   or    0.49^f = 72.

Taking log of both side,

f * log(0.49) = log 72

or f = log 72/log(0.49)

or   f = -5.99518

Therefore,   j^-(144) =-5.996168                           Ans