Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $f(x)=\{\begin{array}{ll}{k}_1(x-\pi {)}^2-1,& x\le \pi \\ k_2\cos x,& x>\pi \end{array}$ is twice differentiable, then the ordered pair $(k_1,k_2)$ is equal to:

Imagine you are an architect designing a bridge that connects two different terrains. On one side, you have a parabolic path, and on the other, a gentle, oscillating cosine wave.

To make this bridge safe for travel, it cannot have any sudden jumps (continuity), it cannot have any sharp, dangerous corners (differentiability), and it must have a smooth transition in its curvature (twice differentiability). This is exactly the challenge we face with our piecewise function:

In calculus, when we say a function is 'twice differentiable,' we are imposing a strict hierarchy of requirements at the junction point $x=\pi$.

First, the function must be continuous, meaning the left-hand limit (LHL) must equal the right-hand limit (RHL). Second, the first derivative $f^{\prime }(x)$ must be continuous, ensuring the slope transitions smoothly.

Third, the second derivative $f^{\prime \prime }(x)$ must be continuous, ensuring the curvature transitions smoothly. Let us peel back these layers one by one.

The Continuity Bridge

For the function to be continuous at $x=\pi$, the two branches must meet at the same point. We calculate the LHL using the first branch:

Now, we calculate the RHL using the second branch:

Equating these, we get $-1=-k_2$, which immediately reveals that $k_2=1$. The first piece of our puzzle is solved!

The Differentiability Mystery

Next, we look at the first derivative. Differentiating the branches, we get $f^{\prime }(x)=2k_1(x-\pi )$ for $x\le \pi$ and $f^{\prime }(x)=-k_2\sin x$ for $x>\pi$.

At $x=\pi$, the left-hand derivative is $2k_1(\pi -\pi )=0$, and the right-hand derivative is $-k_2\sin \pi =0$.

Since $0=0$, the function is differentiable at the junction for any value of $k_1$. This is a fascinating result—the slope is zero on both sides, meaning the transition is perfectly flat regardless of $k_1$.

The Second Derivative and the Final Constant

Finally, we reach the requirement of being twice differentiable. We differentiate $f^{\prime }(x)$ to find $f^{\prime \prime }(x)$.

For the first branch, the derivative of $2k_1(x-\pi )$ is simply $2k_1$. For the second branch, the derivative of $-k_2\sin x$ is $-k_2\cos x$.

For $f^{\prime \prime }(x)$ to be continuous at $x=\pi$, we must have ${\lim }_{x\to {\pi }^{-}}{f}^{\prime \prime }(x)={\lim }_{x\to {\pi }^{+}}{f}^{\prime \prime }(x)$. Substituting $\pi$, we get:

Since $\cos \pi =-1$, this simplifies to $2k_1=-k_2(-1)$, or $2k_1=k_2$. We already know $k_2=1$, so $2k_1=1$, which gives us $k_1=\frac{1}{2}$.

By systematically applying the conditions of continuity, differentiability, and twice differentiability, we have determined that $k_1=\frac{1}{2}$ and $k_2=1$. The ordered pair $(k_1,k_2)$ is $(\frac{1}{2},1)$.

This problem is a beautiful reminder that calculus is not just about crunching numbers; it is about understanding the geometric soul of a function. When we ensure these derivatives match, we are essentially ensuring that the transition between two different mathematical worlds is seamless, elegant, and perfectly smooth.