The inverse of the function f(x) = 1/3x^2 - 3x + 5 is f-1(x) = 9/2 + √[3(x + 7/4)], the sum of the arithmetic series is 1078 and the common ratio of the sequence is 2
<h3>The inverse of the function?</h3>
The function is given as:
f(x) = 1/3x^2 - 3x + 5
Next, we rewrite the function as in vertex form
Using a graphing calculator, the vertex form of the function f(x) = 1/3x^2 - 3x + 5 is
f(x) = 1/3(x - 9/2)^2 - 7/4
Express f(x) as y
y = 1/3(x - 9/2)^2 - 7/4
Swap x and y
x = 1/3(y - 9/2)^2 - 7/4
Add 7/4 to both sides
1/3(y - 9/2)^2 = x + 7/4
Multiply through by 3
(y - 9/2)^2 = 3(x + 7/4)
Take the square root of both sides
y - 9/2 = √[3(x + 7/4)]
Add 9/2 to both sides
y = 9/2 + √[3(x + 7/4)]
Rewrite as an inverse function
f-1(x) = 9/2 + √[3(x + 7/4)]
<h3>Sum of arithmetic series</h3>
Here, we have:
5 + 18 + 31 + 44 + ... 161
Calculate the number of terms using:
L = a + (n -1)d
So, we have:
161 = 5 + (n - 1) * 13
This gives
(n - 1) * 13 = 156
Divide by 13
n - 1 = 12
Add 1
n = 13
The sum is then calculated as:
Sn = n/2 * [a + L]
This gives
Sn =13/2 * (5 + 161)
Evaluate
Sn = 1078
Hence, the sum of the arithmetic series is 1078
<h3>The common ratio of the sequence</h3>
Here, we have:
T11 = 32 * T6
The nth term of a geometric sequence is
Tn = ar^(n-1)
This gives
ar^10 = 32 * ar^5
Divide by ar^5
r^5 = 32
Take the fifth root
r = 2
Hence, the common ratio of the sequence is 2
Read more about sequence at:
brainly.com/question/7882626
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