For a given function f(x) we define the domain restrictions as values of x that we can not use in our function. Also, for a function f(x) we define the inverse g(x) as a function such that:
g(f(x)) = x = f(g(x))
<u>The restriction is:</u>
x ≠ 4
<u>The inverse is:</u>
Here our function is:
We know that we can not divide by zero, so the only restriction in this function will be the one that makes the denominator equal to zero.
(x - 4)^2 = 0
x - 4 = 0
x = 4
So the only value of x that we need to remove from the domain is x = 4.
To find the inverse we try with the general form:
Evaluating this in our function we get:
Remember that the thing above must be equal to x, so we get:
From the two above equations we find:
b = 11
a = 4
Thus the inverse equation is:
If you want to learn more, you can read:
brainly.com/question/10300045
Step-by-step explanation:
step 1. the cube root is also to the 1/3 power
step 2. (27a^3b^7)^(1/3) = (3*3*3*a*a*a*b*b*b*b*b*b*b)^(1/3)
step 3. a third root: 3 factors in the root comes "out" of the root as 1
step 4. 3ab*b(b)^(1/3) = 3ab^2(b)^(1/3).
If a = first term and r = common ratio we have
a + ar + ar^2 = 13 and ar^2 / a = r^2 = 9
so r = 3
and a + 3a + 9a = 13
so a = 1
so they are 1,3 and 9
2.
in geometric series we have
4 , 4r ,4r^2 , 60
Arithmetic;
4, 4r , 4r + d , 4r + 2d
so we have the system of equations
4r + 2d = 60
4r^2 = 4r + d
From first equation
2r + d = 30
so d = 30 - 2r
Substitute for d in second equation:-
4r^2 - 4r - (30-2r) = 0
4r^2 - 2r - 30 =0
2r^2 - r - 15 = 0
(r - 3)(2r + 5) = 0
r = 3 or -2.5
r must be positive so its = 3
and d = 30 - 2(3) = 24
and the numbers are 4*3 = 12 , 4*3^2 = 36
first 3 are 4 , 12 and 36 ( in geometric)
and last 3 are 12, 36 and 60 ( in arithmetic)
The 2 numbers we ause are 12 and 36.
Answer: No, it can never be greater. At most it can be equal though.
