Consider the following for the next two (02) items that follow Let f(x) = \(\left \{ \begin{matrix} x + 6, x \le 1 \\\ px + q, 1 < x < 2 \\\ 5x , x \ge 2 \end{matrix} \right.\) and f(x) is continuous
What is the value of p ?
3
The question asks us to find the value of \(p\) for a given piecewise function \(f(x)\) such that the function is continuous. A function is continuous at a point if the limit of the function as it approaches that point exists and is equal to the function's value at that point. For a piecewise function, we need to check the points where the definition of the function changes.
Understanding Continuity for a Piecewise Function
A piecewise function is continuous over an interval if it is continuous at every point in that interval. For the function given, the definition changes at \(x = 1\) and \(x = 2\). For the function to be continuous everywhere, it must be continuous at these transition points.
The condition for continuity at a point \(c\) is:
- \(f(c)\) is defined.
- \(\lim_{x \to c} f(x)\) exists. This means \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)\).
- \(\lim_{x \to c} f(x) = f(c)\).
We need to apply these conditions at \(x = 1\) and \(x = 2\).
Analyzing Continuity at \(x = 1\)
At \(x = 1\), the function definition changes from \(x + 6\) to \(px + q\). For continuity at \(x=1\), the left-hand limit, the right-hand limit, and the function value must all be equal.
- Function value at \(x = 1\): For \(x \le 1\), \(f(x) = x + 6\). So, \(f(1) = 1 + 6 = 7\).
- Left-hand limit at \(x = 1\): For \(x \le 1\), \(f(x) = x + 6\). So, \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 6) = 1 + 6 = 7\).
- Right-hand limit at \(x = 1\): For \(x > 1\) (specifically \(1 < x < 2\)), \(f(x) = px + q\). So, \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (px + q) = p(1) + q = p + q\).
For continuity at \(x = 1\), we must have \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)\). This gives us the equation:
\(7 = p + q\) (Equation 1)
Analyzing Continuity at \(x = 2\)
At \(x = 2\), the function definition changes from \(px + q\) to \(5x\). For continuity at \(x=2\), the left-hand limit, the right-hand limit, and the function value must all be equal.
- Function value at \(x = 2\): For \(x \ge 2\), \(f(x) = 5x\). So, \(f(2) = 5(2) = 10\).
- Left-hand limit at \(x = 2\): For \(x < 2\) (specifically \(1 < x < 2\)), \(f(x) = px + q\). So, \(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (px + q) = p(2) + q = 2p + q\).
- Right-hand limit at \(x = 2\): For \(x \ge 2\), \(f(x) = 5x\). So, \(\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (5x) = 5(2) = 10\).
For continuity at \(x = 2\), we must have \(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)\). This gives us the equation:
\(2p + q = 10\) (Equation 2)
Solving the System of Equations
We now have a system of two linear equations with two variables, \(p\) and \(q\):
Equation 1: \(p + q = 7\)
Equation 2: \(2p + q = 10\)
We can solve this system to find the values of \(p\) and \(q\). The easiest way to solve for \(p\) is to subtract Equation 1 from Equation 2:
\((2p + q) - (p + q) = 10 - 7\)
\(2p + q - p - q = 3\)
\(p = 3\)
We can also find \(q\) by substituting the value of \(p\) (which is 3) back into Equation 1:
\(3 + q = 7\)
\(q = 7 - 3\)
\(q = 4\)
Determining the Value of p
The question asks for the value of \(p\). From our calculation, we found that \(p = 3\).
Thus, for the piecewise function to be continuous, the value of \(p\) must be 3.
| Point | Condition for Continuity | Equation from \(f(x)\) |
|---|---|---|
| \(x=1\) | \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)\) | \(1+6 = p(1)+q \implies 7 = p+q\) |
| \(x=2\) | \(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)\) | \(p(2)+q = 5(2) \implies 2p+q = 10\) |
Solving the system:
\(2p + q = 10\)
\(p + q = 7\)
Subtracting the second equation from the first:
\((2p + q) - (p + q) = 10 - 7\)
\(p = 3\)
Conclusion on Piecewise Continuity
By ensuring the limits match at the points where the function definition changes (\(x=1\) and \(x=2\)), we derived a system of equations for \(p\) and \(q\). Solving this system gave us \(p=3\) and \(q=4\). Therefore, the value of \(p\) that makes the function continuous is 3.
Revision Table: Key Continuity Concepts
| Concept | Description | Condition |
|---|---|---|
| Continuity at a Point \(c\) | The graph has no breaks, jumps, or holes at \(c\). | \(\lim_{x \to c} f(x) = f(c)\) |
| Left-Hand Limit | The limit as \(x\) approaches \(c\) from values less than \(c\). | \(\lim_{x \to c^-} f(x)\) |
| Right-Hand Limit | The limit as \(x\) approaches \(c\) from values greater than \(c\). | \(\lim_{x \to c^+} f(x)\) |
| Limit Existence | The limit at \(c\) exists if the left and right limits are equal. | \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)\) |
Additional Information: Solving Systems for Continuous Functions
Problems involving continuous piecewise functions often require solving systems of linear equations. Each point where the function definition changes provides an equation based on the continuity condition. If there are \(n\) such points and \(m\) unknown parameters (like \(p\) and \(q\) in this case), you typically need at least \(m\) independent equations to find unique values for the parameters. In this problem, we had two transition points (\(x=1\) and \(x=2\)) and two unknown parameters (\(p\) and \(q\)), leading to a solvable system of two equations.
The process involves:
- Identify transition points in the piecewise function.
- At each transition point \(c\), set the left-hand limit equal to the right-hand limit (and the function value if needed, though LHL=RHL usually suffices if the function is defined appropriately at the point).
- Formulate a system of equations based on these equalities.
- Solve the system for the unknown parameters.
Similar Questions
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