If the derivative of the function f(x)= array cc b x^2+a x+4 ; & x -1 a x^2+b ; & x < -1 array . is everywhere continuous, then

Step-by-step solution

The function f(x) is defined as f(x) = bx^2 + ax + 4 for x -1 and f(x) = ax^2 + b for x For the derivative f'(x) to be continuous everywhere, f(x) must be continuous at x = -1 . Thus, _ x -1⁻ f(x) = _ x -1⁺ f(x) = f(-1) . Calculating the limits: _ x -1⁻ (ax^2 + b) = a + b and _ x -1⁺ (bx^2 + ax + 4) = b - a + 4 . Equating these, a + b = b - a + 4 , which simplifies to 2a = 4 , so a = 2 . Now, consider the derivative f'(x) = array cc 2bx + a; & x > -1 2ax; & x For f'(x) to be continuous at x = -1 , the left-hand der