We have the function:
f(x) = 3x / (x + 7)
(a)
We rename the function as: f(x) = y
Then:
y = 3x / (x + 7)
Taking the inverse:
1/y = (x + 7) / 3x
1/y = x/3x + 7/3x
1/y = 1/3 + 7/3x
Solving for x:
1/y - 1/3 = 7/3x
1/x = 3/7y - 1/7 = (3 - y) / 7y
Taking the inverse:
x = 7y / (3 - y)
Then, the inverse function of f is:
f ⁻¹(x) = 7x / (3 - x)
(b)
We know that the division by 0 is undefined in real numbers. From the function f, we have a division by 0 if x = -7, so the domain should be:
Dom_f = {x| x ≠ -7}
For the range, we know that x = -7 is a vertical asymptote of the function f, so this means that the graph never passes across x = -7, but it tends to it on infinity. Then, the range of f is:
Ran_f = All the real numbers
For f ⁻¹(x), we see that for x = 3 there is a division by 0, so this is an asymptote of the function. Then, the domain of f ⁻¹ is:
Dom_f ⁻¹ = {x| x ≠ 3}
Again, as there is an asymptote, the range is:
Ran_f ⁻¹ = All the real numbers
Answer:
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Step-by-step explanation:
Answer:
9 fluid ounces of sparkling soda.
Step-by-step explanation:
The ratio is 1 to 8 so:
72 / 8 = 9
Order your terms by the powers of exponents in decreasing order (as is the case with pure number division hundreds, tens, ones, tenths, etc, etc) as x^n, x^n-1 etc...
(-x^4+1)/(x-1)
-x^3 rem -x^3-1
-x^2 rem -x^2-1
-x rem -x-1
-1 rem 0
(x-1)(-x^3-x^2-x-1)
(x-1)(-x^3-x-x^2-1)
(x-1)(-x(x^2+1)-1(x^2+1))
(x-1)(-x-1)(x^2+1)
