Understanding Continuity of Piecewise Functions

A function $f(x)$ is said to be continuous at a point $x = a$ if the following three conditions are met:

1. The function $f(a)$ is defined.
2. The limit of the function as $x$ approaches $a$ exists, i.e., $\lim_{x \to a} f(x)$ exists. This requires that the left-hand limit and the right-hand limit are equal: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$.
3. The limit of the function at $x = a$ is equal to the function value at $x = a$, i.e., $\lim_{x \to a} f(x) = f(a)$.

For a piecewise function like the one given, continuity at the point where the definition changes (in this case, $x = \pi$) is crucial. We need to ensure that the two pieces meet seamlessly at this point.

Analyzing the Given Piecewise Function

The function is defined as:

\[ f(x) = \begin{cases} kx + 1 & \text{if } x \le \pi \\ \cos x & \text{if } x > \pi \end{cases} \]

We are told that $f(x)$ is continuous at $x = \pi$. We will use the conditions for continuity at $x = \pi$ to find the value of $k$.

Step-by-Step Solution for Continuity at \(x = \pi\)

Step 1: Evaluate the function at \(x = \pi\)

Since the function is defined as $f(x) = kx + 1$ for $x \le \pi$, we use this part to find $f(\pi)$.

\[ f(\pi) = k(\pi) + 1 \] \[ f(\pi) = k\pi + 1 \]

Step 2: Evaluate the left-hand limit at \(x = \pi\)

The left-hand limit as $x$ approaches $\pi$ (from values less than or equal to $\pi$) uses the definition $f(x) = kx + 1$.

\[ \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx + 1) \]

Since $kx + 1$ is a polynomial function, the limit can be found by direct substitution.

\[ \lim_{x \to \pi^-} (kx + 1) = k(\pi) + 1 \] \[ \lim_{x \to \pi^-} f(x) = k\pi + 1 \]

Step 3: Evaluate the right-hand limit at \(x = \pi\)

The right-hand limit as $x$ approaches $\pi$ (from values greater than $\pi$) uses the definition $f(x) = \cos x$.

\[ \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (\cos x) \]

Since $\cos x$ is a continuous function, the limit can be found by direct substitution.

\[ \lim_{x \to \pi^+} (\cos x) = \cos(\pi) \]

We know that $\cos(\pi) = -1$.

\[ \lim_{x \to \pi^+} f(x) = -1 \]

Step 4: Apply the Continuity Condition

For the function to be continuous at $x = \pi$, the left-hand limit, the right-hand limit, and the function value at $x = \pi$ must all be equal.

\[ \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^+} f(x) = f(\pi) \]

From our calculations, we have:

\[ k\pi + 1 = -1 = k\pi + 1 \]

We can use the equation formed by equating the left-hand limit (or function value) and the right-hand limit:

\[ k\pi + 1 = -1 \]

Step 5: Solve for the value of \(k\)

Now, we solve the equation $k\pi + 1 = -1$ for $k$.

Subtract 1 from both sides:

\[ k\pi = -1 - 1 \] \[ k\pi = -2 \]

Divide by $\pi$ (since $\pi \ne 0$):

\[ k = \frac{-2}{\pi} \]

Conclusion

For the function $f(x)$ to be continuous at $x = \pi$, the value of $k$ must be $-\frac{2}{\pi}$.

Comparing with Options

Let's compare our calculated value of $k$ with the given options:

1. $0$
2. $\pi$
3. $2 / \pi$
4. $- 2 / \pi$

Our calculated value $k = -\frac{2}{\pi}$ matches option 4.

Revision Table: Key Continuity Concepts

Additional Information on Function Continuity

Understanding continuity is fundamental in calculus. Here are some related points:

• Types of Discontinuity: If a function is not continuous at a point, it is said to have a discontinuity. Common types include removable discontinuity (where the limit exists but is not equal to the function value, or the function isn't defined), jump discontinuity (where the left and right limits exist but are not equal), and essential discontinuity (where at least one limit does not exist, e.g., infinite discontinuity).
• Continuity on an Interval: A function is continuous on an open interval $(a, b)$ if it is continuous at every point in the interval. It is continuous on a closed interval $[a, b]$ if it is continuous on $(a, b)$, continuous from the right at $a$ ($\lim_{x \to a^+} f(x) = f(a)$), and continuous from the left at $b$ ($\lim_{x \to b^-} f(x) = f(b)$).
• Properties of Continuous Functions: Sums, differences, products, and quotients (where the denominator is non-zero) of continuous functions are continuous. Compositions of continuous functions are also continuous.
• Intermediate Value Theorem (IVT): A key theorem for continuous functions. If $f$ is continuous on the closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in $(a, b)$ such that $f(c) = N$.