(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

2 Answers
Feb 4, 2018

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

Feb 4, 2018

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