It's C. When you're dividing g(x) by f(x), you get the equations in both B. and C., but C is more correct because you can't have x=-1/3 because the function is undefined at that number (you can't have the denominator equal 0)
Answer:C
Step-by-step explanation: When a table represents a nonlinear function, the rate of change is not constant. A wouldn't be the answer because the rate of change is always +10 (you would add 10 to get from -9 to 1; you would add 10 to get from 1 to 10) . It wouldn't be B because the rate of change is always -2, and the rate of change for D is always +3. For C, however, the rate of change is not constant all the way through (to get from 0 to 6, you would add 6, but to get from 6 to 16 you add 10).
When you have the same denominator. Other than that have different denominators will not be possible.
Answer:
A.) gf(x) = 3x^2 + 12x + 9
B.) g'(x) = 2
Step-by-step explanation:
A.) The two given functions are:
f(x) = (x + 2)^2 and g(x) = 3(x - 1)
Open the bracket of the two functions
f(x) = (x + 2)^2
f(x) = x^2 + 2x + 2x + 4
f(x) = x^2 + 4x + 4
and
g(x) = 3(x - 1)
g(x) = 3x - 3
To find gf(x), substitute f(x) for x in g(x)
gf(x) = 3( x^2 + 4x + 4 ) - 3
gf(x) = 3x^2 + 12x + 12 - 3
gf(x) = 3x^2 + 12x + 9
Where
a = 3, b = 12, c = 9
B.) To find g '(12), you must first find the inverse function of g(x) that is g'(x)
To find g'(x), let g(x) be equal to y. Then, interchange y and x for each other and make y the subject of formula
Y = 3x + 3
X = 3y + 3
Make y the subject of formula
3y = x - 3
Y = x/3 - 3/3
Y = x/3 - 1
Therefore, g'(x) = x/3 - 1
For g'(12), substitute 12 for x in g' (x)
g'(x) = 12/4 - 1
g'(x) = 3 - 1
g'(x) = 2.
