Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f(x)=\{\begin{array}{ll}\frac{1}{|x|},& |x|\ge 1\\ a{x}^2+b,& |x|<1\end{array}$ is differentiable at every point of the domain, then the values of $a$ and $b$ are respectively:

We are given the piecewise function:

The outer branches, defined for $|x|\ge 1$, represent the fixed hyperbola $y=\frac{1}{|x|}$. The inner segment, defined for $|x|<1$, is a parabola $y=a{x}^2+b$. Our objective is to determine the constants $a$ and $b$ such that the function is differentiable at the transition points $x=1$ and $x=-1$.

Ensuring Continuity

Before addressing smoothness, we must ensure the tracks meet at the transition points. This is the condition of continuity.

At $x=1$, the limit of the parabola must equal the limit of the hyperbola:

This simplifies to our first fundamental equation:

This equation guarantees that there is no physical gap in the graph at the junction.

Matching the Slopes

To prevent a sharp corner, the slopes of the two functions must be identical at the point of contact. This is the condition of differentiability.

We calculate the derivatives of both segments: 1. The derivative of the parabola $a{x}^2+b$ is $2ax$. 2. The derivative of the hyperbola $\frac{1}{x}$ (for $x>0$) is $-\frac{1}{x^2}$.

Evaluating these derivatives at the transition point $x=1$:

Setting these equal provides our second equation:

We now possess a system of two linear equations:

1. $a+b=1$ 2. $2a=-1$

From the second equation, we immediately find:

Substituting this value into the first equation:

By applying the conditions of continuity and differentiability, we have determined the unique values: $a=-\frac{1}{2}$ and $b=\frac{3}{2}$. The resulting parabola, $y=-\frac{1}{2}{x}^2+\frac{3}{2}$, creates a perfectly smooth transition for the hyperbola.