Tamika finds an expression for S(r) that gives the surface area in square inches of a cylindrical can with a specific fixed volume in terms of its radius r in inches. Shown is the graph Tamika gets if she allows r to take on any value between -1 and 5. Part A: What would be a more appropriate domain for Tamika to use instead?

Part B: What is the approximate minimum surface area for the can?

Part C: Consider the intervals shown.
I. 16 < y < 60

II. 0 < x < ∞

III. (1, ∞)

IV. (-60, 0)

V. (2, 3)
Which interval for the function S(r) represents an increase in the surface area?

A.

I and IV
B.

III and V
C.

I, II, and IV
D.

II, III, and V

3

Step-by-step explanation:

answer: 298143

step-by-step explanation:

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2