The option that is true with regard to the following functions is Option B. "The domain g(x) and h(x) include all real number while the domain of i(x) and h(x) are restricted"
<h3>What is the explanation for the above?</h3>
- Lets examine f(x) = 3x + 14
Note that this function is indicative of a straight-line. See the attached graph for function 1. Note that it doesn't have any end points. That is, it is Asymptote.
- Let us examine h(x) = 3ˣ + 1
This represents an exponential graph. Just like the function above it doesn't have any end point. It however has an asymptote:
y = 0
- Let us look at F'(x) = Log₃ (x = 1).
This is indicative of logarithm graph. It doesn't have any end but has an asymptote x == 0
- Let us take a look at g(x) = X⁴ + 3x² - 14
Notice that in the mid point there is an end point given as (0, -14). Thus, it is correct to state that the function in Option B is the only one that exhibits end behavior and as such is restricted.
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The domain is all possible x values.
In this case it goes from -2 to 6
So your domain is [-2,6] which is option b
Since the values -2 and 6 are included you use greater than or equal to and less than or equal to.
I
The two curves meet in the first quadrant when

Then the integral in question is

Answer:
(-10; 60)
Step-by-step explanation:
y = -6x
(12; -2)
-2 = -6 · 12
-2 = -72 - no
(3; -9)
-9 = -6 · 3
-9 = -18 - no
(-10; 60)
60 = -6 · (-10)
60 = 60 - yes
(18; -3)
-3 = -6 · 18
-3 = -78 - no