If Rolle's theorem is satisfied for the following function x^3 - ax^2 + bx - 4, x in [1,2] with f'(4/3) = 0, then the ordered pair (a, b) is equal to.

Steps to Solve

1. Identify the function and its derivative We start with the function given by ( f(x) = x^3 - ax^2 + bx - 4 ).

2. Find the derivative of the function The derivative ( f'(x) ) is calculated as follows:

[ f'(x) = 3x^2 - 2ax + b ]

3. Apply the condition f'(4/3) = 0 Substituting ( x = \frac{4}{3} ):

[ f'\left(\frac{4}{3}\right) = 3\left(\frac{4}{3}\right)^2 - 2a\left(\frac{4}{3}\right) + b = 0 ]

Calculating ( \left(\frac{4}{3}\right)^2 = \frac{16}{9} ):

[ f'\left(\frac{4}{3}\right) = 3 \cdot \frac{16}{9} - \frac{8a}{3} + b = 0 ]

This simplifies to:

[ \frac{48}{9} - \frac{8a}{3} + b = 0 ]

Multiply the entire equation by 9 to eliminate the fraction:

[ 48 - 24a + 9b = 0 \quad (1) ]

4. Use Rolle's theorem conditions According to Rolle's theorem, since ( f(1) = f(2) ):

Calculate ( f(1) ):

[ f(1) = 1^3 - a \cdot 1^2 + b \cdot 1 - 4 = 1 - a + b - 4 = -a + b - 3 \quad (2) ]

Calculate ( f(2) ):

[ f(2) = 2^3 - a \cdot 2^2 + b \cdot 2 - 4 = 8 - 4a + 2b - 4 = -4a + 2b + 4 \quad (3) ]

Setting equations (2) and (3) equal to each other:

[ -a + b - 3 = -4a + 2b + 4 ]

Rearranging gives:

[ 3a - b - 7 = 0 \quad (4) ]

5. Solve the system of equations Now we solve equations (1) and (4) simultaneously:

1. ( -24a + 9b = -48 ) (from equation 1)

2. ( 3a - b = 7 ) (from equation 4)

From equation (4), express ( b ):

[ b = 3a - 7 ]

Substituting into equation (1):

[ -24a + 9(3a - 7) = -48 ]

Distributing:

[ -24a + 27a - 63 = -48 ]

Combining like terms:

[ 3a - 63 = -48 \implies 3a = 15 \implies a = 5 ]

Substituting ( a ) back to find ( b ):

[ b = 3(5) - 7 = 15 - 7 = 8 ]

Thus, the ordered pair ( (a, b) ) is ( (5, 8) ).