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