Rolle's Theorem is valid for the given function f(x)=x3+bx2+cx,1≤x≤2 at the point x= 43, then values of b and c are respectively

Rolle's Theorem states that if a function f(x) is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one c in the open interval (a, b) such that f'(c) = 0.

Given the function f(x) = x^3 + bx^2 + cx, we need to find the values of b and c such that the theorem is valid at the point x = 4/3.

Step 1: Check if the function is continuous and differentiable on the interval [1, 2]. Since f(x) is a polynomial function, it is continuous and differentiable on the entire real line, including the interval [1, 2].

Step 2: Check if f(1) = f(2). Substituting x = 1 and x = 2 into the function, we get:

f(1) = 1 + b + c f(2) = 8 + 4b + 2c

For Rolle's Theorem to be valid, f(1) must equal f(2), so we have the equation:

1 + b + c = 8 + 4b + 2c

Solving this equation gives us:

3b + c = 7

Step 3: Find f'(x) and set it equal to zero. The derivative of f(x) is:

f'(x) = 3x^2 + 2bx + c

Setting f'(x) = 0 gives us the equation:

3x^2 + 2bx + c = 0

Substituting x = 4/3 into this equation gives us:

16/3 + 8b/3 + c = 0

Solving this equation gives us:

8b + 3c = -16

Step 4: Solve the system of equations. We now have a system of two equations in two variables:

3b + c = 7 8b + 3c = -16

Solving this system gives us the values of b and c.