Given the function f(x) = 4(x+3) − 5, solve for the inverse function when x = 3. (1 point)
2 answers:
the function f(x) = 4(x+3) − 5
Lets find the inverse function f^-1(x)
step 1: Replace f(x) with y
y = 4(x+3) − 5
step 2: Replace x with y and y with x
x = 4(y+3) - 5
step 3: Solve for y
x = 4(y+3) - 5
x = 4y +12 -5
x= 4y + 7 (subtract 7 on both sides)
x - 7 = 4y (divide by 4 on both sides)
Given : x = 3
We plug in 3 for x in the inverse function
= -1
the inverse function when x = 3 is -1
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The formula for the nth term of a geometric sequence:
a₁ - the first term, r - the common ratio
![54, a_2, a_3, 128 \\ \\ a_1=54 \\ a_4=128 \\ \\ a_n=a_1 \times r^{n-1} \\ a_4=a_1 \times r^3 \\ 128=54 \times r^3 \\ \frac{128}{54}=r^3 \\ \frac{128 \div 2}{54 \div 2}=r^3 \\ \frac{64}{27}=r^3 \\ \sqrt[3]{\frac{64}{27}}=\sqrt[3]{r^3} \\ \frac{\sqrt[3]{64}}{\sqrt[3]{27}}=r \\ r=\frac{4}{3}](https://tex.z-dn.net/?f=54%2C%20a_2%2C%20a_3%2C%20128%20%5C%5C%20%5C%5C%0Aa_1%3D54%20%5C%5C%0Aa_4%3D128%20%5C%5C%20%5C%5C%0Aa_n%3Da_1%20%5Ctimes%20r%5E%7Bn-1%7D%20%5C%5C%0Aa_4%3Da_1%20%5Ctimes%20r%5E3%20%5C%5C%0A128%3D54%20%5Ctimes%20r%5E3%20%5C%5C%0A%5Cfrac%7B128%7D%7B54%7D%3Dr%5E3%20%5C%5C%20%5Cfrac%7B128%20%5Cdiv%202%7D%7B54%20%5Cdiv%202%7D%3Dr%5E3%20%5C%5C%0A%5Cfrac%7B64%7D%7B27%7D%3Dr%5E3%20%5C%5C%0A%5Csqrt%5B3%5D%7B%5Cfrac%7B64%7D%7B27%7D%7D%3D%5Csqrt%5B3%5D%7Br%5E3%7D%20%5C%5C%0A%5Cfrac%7B%5Csqrt%5B3%5D%7B64%7D%7D%7B%5Csqrt%5B3%5D%7B27%7D%7D%3Dr%20%5C%5C%0Ar%3D%5Cfrac%7B4%7D%7B3%7D)
a₁ - the first term, r - the common ratio
Answer:
Step-by-step explanation:
Lets divide it in cases, then sum everything
Case (1): All 5 numbers are different
In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.
The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.
Case (2): 4 numbers are different
We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4
We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.
The total cardinality of this case is
Case (3): 3 numbers are different
As we did before, we pick 3 elements from a set of n. The amount of possibilities is
Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of
Case (4): 2 numbers are different
We pick 2 numbers from a set of n, with a total of possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.
The total amount of possibilities for this case is
Case (5): All numbers are the same
This is easy, he have as many possibilities as numbers the set has. In other words, n
Conclussion
By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is
I hope that works for you!