Question

(The function is cubed) How do you find a and b?

`f(x)=3(ax-b/x)^3`
Given `f` `(3/2)` = `3`
and `f'` `(3/2)` = 30

Answer 1

`a=2;b=3`

Explanation:

Since
`f(3/2)=3(3/2a-b/(3/2))^3`
we get:
`cancel3(3/2a-b/(3/2))^3=cancel3^1`
that's

`color(red)(3/2a-2/3b=1)`

Since
`f'(x)=3*3(ax-b/x)^2*(a+b/x^2)`
we get
`f'(3/2)=9(3/2a-b/(3/2))^2*(a+b/(3/2)^2)=30`
that's
`cancel9^3(3/2a-2/3b)^2*(a+4/9b)=cancel30^10`

Let's substitute `color(red)(3/2a-2/3b=1)`, then we get:

`3(a+4/9b)=10`
that's

`color(blue)(3a+4/3b=10)`

Since `color(red)(3/2a-2/3b=1)->3a=2+4/3b`
and we'll substitute it in `color(blue)(3a+4/3b=10)`
The we get:

`4/3b+4/3b+2=10`
`8/3b-8=0`
`b=3`

Since `a=2/3+4/9b`, we get:
`b=3->a=2/3+4/3=2`

Answer 2

Please see below.

Explanation:

`f(x)=3(ax-b/x)^3`

We are given: `f(3/2) = 3`

Using the definition of `f`, we get
`f(3/2)=3(3/2a-b/(3/2))^3 = 3`

So `3/2a-2/3b = 1`.

Clear fractions by multiplying by `6` to get

`color(red)(9a-4b=6)`

We are also given some information about the derivative of `f`, so
we find the derivative:

`f'(x) = 9(ax-b/x)^2(a+b/x^2)`

Using `x = 3/2`, gives us

`f'(3/2) = 9(3/2a-2/3b)^2(a+b/(3/2)^2)`

`= 9(3/2a-2/3b)^2(a+4/9b)`

`= (3/2a-2/3b)^2(9a+4b)`

Recall from above that the first factor is `1` and also we were given that `f'(3/2) = 30`. So,

`f'(3/2) = (1)^2(9a+4b) = 30` so

`color(blue)(9a+4b=30)`

Solving the system

`9a+4b=30`
`9a-4b=6`

We get

`18a=36` so `a = 2` and `8b=24` so `b = 3`