Question

For what value of \(k\) is the function \(g(x)=\left\\{\begin{array}{ll}x^{2}+5, & x \leq 3 \\ 2 x-k, & x>3\end{array}\right.\) continuous at \(x=3\) ?

Short answer

Solution: Based on the analysis and calculations, the value of \(k\) that makes the function continuous at \(x=3\) is \(-8\).

Step-by-step solution

Understand the definition of continuity

A function is continuous at a point if the following three conditions are met: 1. The function is defined at the point. 2. The limit of the function exists at the point. 3. The limit of the function is equal to the function's value at the point. We will use this definition to find the value of \(k\) that makes the function continuous at \(x=3\).

Find the function's value at \(x=3\)

We will first find the function's value at \(x=3\) when \(x\leq 3\). To do this, we will use the first part of our piecewise function: \(g(x) = x^2 + 5\). Evaluate the function at \(x=3\): \(g(3) = 3^2 + 5 = 9 + 5 = 14\). Now, we will calculate the function's value at \(x=3\) using the second part of our piecewise function, which is applicable when \(x>3\): \(g(x) = 2x - k\). Since we want the function to be continuous at \(x=3\), it must be equal to the value we found in the previous step. Therefore, we can set the two values equal: \(2(3) - k = 14\).

Solve for \(k\)

Now we can solve the equation \(2(3) - k = 14\) to find the value of \(k\) that makes the function continuous at \(x=3\): \(6 - k = 14\) Add \(k\) to both sides: \(6 = k + 14\) Subtract \(14\) from both sides: \(k = -8\) So the value of \(k\) that makes the function continuous at \(x=3\) is \(-8\).

Key concepts

Piecewise Function

A piecewise function is a function defined by multiple sub-functions. Each sub-function applies to a specific interval of the main function's domain. In the case of the function given, it comprises two distinct rules based on the value of \(x\):

• For \(x \leq 3\), the function is described by \(x^2 + 5\).
• For \(x > 3\), the function is given by \(2x - k\).

This structuring allows for flexibility as it can model complex situations where a single expression cannot capture all possible scenarios. By understanding each condition or branch, one can decipher how the function behaves throughout its domain. Remember, determining the core factors of where one rule stops and another begins is crucial in piecing together the complete function.

Continuity in Functions

A function is said to be continuous at a certain point if you can plot it without lifting your pen from the paper. Mathematically, for a function to be continuous at a point \(x = a\), three conditions must be satisfied:

• The function must be defined at \(a\).
• The limit of the function as \(x\) approaches \(a\) must exist.
• The value of the function at \(a\) must equal the limit as \(x\) approaches \(a\).

In the given exercise, these principles were applied at \(x = 3\). Since the function's value must be equal from both the left and the right as it meets at this point, \(k\) needs adjustment so both rules give the same value when \(x = 3\). By equating \(g(3) = 14\) from both expressions, we ensure the continuity of the function at \(x = 3\).

Limit of a Function

The limit of a function at a given point describes the value that the function approaches as \(x\) gets arbitrarily close to that point. In simple terms, if you hone in on a point from either side on a graph, the limit tells you which value you are steering toward.
In our problem, we are assessing the limit as \(x\) approaches \(3\). From the left (\(x \leq 3\)), the expression \(x^2 + 5\) simplifies to \(14\) when \(x = 3\). From the right (\(x > 3\)), by setting the expression \(2x - k\) equal to \(14\), we find that continuity holds only if the expected value matches: \(2 \cdot 3 - k = 14\).
To solve for \(k\), these limits are explored in both directions to confirm they converge at the specified point. This tidy convergence and equating ensure the function behaves similarly approaching from both sides, thus confirming a smooth transition at \(x = 3\).