How do I figure it out the value of constants a≥0a≥0, bb and cc that will be the function f(x) = \begin{dcases*} \dfrac{\sin(1 - ax)}{x+1}, & x < -1 \\ \dfrac{\sqrt{x^{4} +11x^2 + 4} - bx^{2} - x - c}{x}, & -1 \leq x < 0 \\ a, & x = 0 \\ \dfrac{\ln(1+ax)}{x}, & x > 0 \end{dcases*}f(x) = \begin{dcases*} \dfrac{\sin(1 - ax)}{x+1}, & x < -1 \\ \dfrac{\sqrt{x^{4} +11x^2 + 4} - bx^{2} - x - c}{x}, & -1 \leq x < 0 \\ a, & x = 0 \\ \dfrac{\ln(1+ax)}{x}, & x > 0 \end{dcases*} continuous everywhere?

 To ensure the function \(f(x)\) is continuous everywhere, we need to check three conditions at the points where the piecewise definition changes:

1. At \(x = -1\), the left-hand limit should equal the right-hand limit.

2. At \(x = 0\), the function value should be equal to the limit from the left and the limit from the right.

3. At \(x = 0\), the left-hand limit should equal the right-hand limit.

Let's go step by step:

1. At \(x = -1\):

   \[

   \lim_{{x \to -1^-}} \frac{\sin(1 - ax)}{x+1} = \lim_{{x \to -1^+}} \left(\frac{\sqrt{x^{4} +11x^2 + 4} - bx^{2} - x - c}{x}\right)

   \]

2. At \(x = 0\):

   \[

   f(0) = a, \quad \lim_{{x \to 0^-}} \frac{\sqrt{x^{4} +11x^2 + 4} - bx^{2} - x - c}{x}, \quad \lim_{{x \to 0^+}} \frac{\ln(1+ax)}{x}

   \]

3. Ensure the left-hand limit at \(x = 0\) equals the right-hand limit:

   \[

   \lim_{{x \to 0^-}} \frac{\sqrt{x^{4} +11x^2 + 4} - bx^{2} - x - c}{x} = \lim_{{x \to 0^+}} \frac{\ln(1+ax)}{x}

   \]

Solving these equations will give you the values of \(a\), \(b\), and \(c\) that make the function continuous everywhere.