SOLUTION: Suppose f(x) = 3(ax-b/x)^3. Given that f(3/2) = 3 and f'(3/2) = 30, find a and b.

Question 1192280: Suppose f(x) = 3(ax-b/x)^3. Given that f(3/2) = 3 and f'(3/2) = 30, find a and b.
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to solve this problem:
**1. Use the given information about f(3/2):**
* f(x) = 3(ax - b/x)³
* f(3/2) = 3(a(3/2) - b/(3/2))³ = 3
* (3a/2 - 2b/3)³ = 1
* 3a/2 - 2b/3 = 1 (Taking the cube root of both sides)
* 9a - 4b = 6 --- (Equation 1)
**2. Find the derivative of f(x):**
* f'(x) = 9(ax - b/x)² * (a + b/x²)
**3. Use the given information about f'(3/2):**
* f'(3/2) = 9(3a/2 - 2b/3)² * (a + 4b/9) = 30
* (3a/2 - 2b/3)² * (a + 4b/9) = 10
* (1)² * (a + 4b/9) = 10 (Since we know from step 1 that 3a/2 - 2b/3 = 1)
* a + 4b/9 = 10
* 9a + 4b = 90 --- (Equation 2)
**4. Solve the system of equations:**
Now we have two equations:
* 9a - 4b = 6 --- (Equation 1)
* 9a + 4b = 90 --- (Equation 2)
Add Equation 1 and Equation 2:
* 18a = 96
* a = 96/18 = 16/3
Substitute the value of 'a' back into Equation 1:
* 9(16/3) - 4b = 6
* 48 - 4b = 6
* 4b = 42
* b = 42/4 = 21/2
**Solution:**
a = 16/3
b = 21/2

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