Find the domain and range of the function. ​f(x)=24+x^2

Domain:

Range:

Domain:

In finding the Domain of a function, the values for the input of the function should not make the output of the function undefined or complex. Because of this, we can think of the values for the input that make the output of the function undefined or complex so that we will not include them in our Domain. We can only make the output undefined if the input makes the denominator . In , there's no value for that makes the denominator as it is constant, (Note: All expressions implicitly have as their denominator even though it's not written). We can only make the output of the function complex if the value of makes the function take the root of a negative number where . There's no radical sign in so we shouldn't worry about the output of being complex. Because there's no value for the input, , that can make the output of undefined or complex, its Domain can be any number.

Domain:

Range:

In finding the Range, it is actually the same logic as finding the Domain but first, we'll have to do a bit of rewriting for the given function.

Let so .

First, we need to make the subject of the equation.

Making the subject:

.

In the Domain, we'll have to think of the value for the input, , that makes the output undefined or complex. In the Range, the same logic for the Domain, we'll have to think for the value of , that makes the undefined or complex. That's why we made in the equation the subject. In our rewritten equation, we can see that is under the square root. Which means if is negative, will be complex. So we have to make be greater than or equal to () so that won't be complex.

Solving for the inequality,

Range:

a ruler? idk

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