A2-column table with 8 rows. the first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2, 3, 4. the second column is labeled f of x with entries negative 2, 0, 2, 2, 0, negative 8, negative 10, negative 20. which could be the entire interval over which the function, f(x), is positive? (–∞, 1) (–2, 1) (–∞, 0) (1, 4)

Step-by-step explanation:

So I'm going to  make the table here.

x     f(x)

-3     -2

-2     0

-1     2

0     2

1     0

2     -8

3     -10

4     -20

Without looking at the answers we are going to deduce what intervals could make f(x) positive.  Well, where  is it shown as positive for the numbers given?  well f(-1) and f(0) are positive, both at 2.  then both f(-2) and f(1) are 0, then everything below -2 and above 1 make f(x) negative, so it looks like everything up until -2 is negative, but increasing then at -2 it turns to 0 and keeps increasing  at least until some time after f(-1).  It might even just stay at 2 until it gets to f(0), and then after that it decreases.  Hopefully that makes sense.

Given the description, and if you on't understand I can try to explain more carefully, where does it look like it's positive?  Well, it goes from negative to 0 at -2, then after that it's positive until 1, where it turns to 0 again, before decreasing.    so everything between -2 and 1 look to make f(x) positive.  Since f(-2) and f(1) don't themselves though it means they are not included, so you use parenthesis.

so the answer is (-2,1)  Again, let me know if you need more of an 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

the equation is not included but the answer is (1.50, 2.00, 2.75)

step-by-step explanation: