If the function \(\rm f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {a + bx,\;\;}&{x < 1}\\ {5,}&{x = 1}\\ {b - ax,}&{x > 1} \end{array}} \right.\) is continuous, then what is the value of (a + b)?
5
Understanding Continuity of a Piecewise Function
A function is considered continuous at a point \(x=c\) if three conditions are met:
- The function value \(f(c)\) exists.
- The limit of the function as \(x\) approaches \(c\) from the left (\(\lim_{x \to c^-} f(x)\)) exists.
- The limit of the function as \(x\) approaches \(c\) from the right (\(\lim_{x \to c^+} f(x)\)) exists.
- The left-hand limit, the right-hand limit, and the function value at \(c\) are all equal: \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)\).
For the given piecewise function \(\rm f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {a + bx,\;\;}&{x < 1}\\ {5,}&{x = 1}\\ {b - ax,}&{x > 1} \end{array}} \right.\) to be continuous, it must be continuous at the point where the definition changes, which is \(x = 1\).
Applying Continuity Conditions at x = 1
Let's check the conditions for continuity at \(x=1\):
- Function Value at x = 1:
The function is defined as \(f(x) = 5\) for \(x=1\).
So, \(f(1) = 5\). This value exists. - Left-Hand Limit at x = 1:
For \(x < 1\), the function is \(f(x) = a + bx\).
The left-hand limit as \(x\) approaches 1 is:
\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (a + bx)\)
Since \(a+bx\) is a polynomial, we can find the limit by direct substitution:
\(\lim_{x \to 1^-} (a + bx) = a + b(1) = a + b\) - Right-Hand Limit at x = 1:
For \(x > 1\), the function is \(f(x) = b - ax\).
The right-hand limit as \(x\) approaches 1 is:
\(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (b - ax)\)
Since \(b-ax\) is a polynomial, we can find the limit by direct substitution:
\(\lim_{x \to 1^+} (b - ax) = b - a(1) = b - a\)
Solving for Constants a and b
For the function to be continuous at \(x=1\), the left-hand limit, the right-hand limit, and the function value must be equal:
\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)\)
Substituting the values we found:
\(a + b = b - a = 5\)
This gives us two equations:
Equation 1: \(a + b = 5\)
Equation 2: \(b - a = 5\)
We can solve this system of linear equations. Let's add the two equations:
\((a + b) + (b - a) = 5 + 5\)
\(a + b + b - a = 10\)
\(2b = 10\)
\(b = \frac{10}{2}\)
\(b = 5\)
Now substitute the value of \(b\) (which is 5) into Equation 1:
\(a + 5 = 5\)
\(a = 5 - 5\)
\(a = 0\)
So, the values of the constants are \(a=0\) and \(b=5\).
Calculating the Value of (a + b)
The question asks for the value of \((a + b)\). Using the values we found for \(a\) and \(b\):
\(a + b = 0 + 5 = 5\)
Thus, the value of \((a + b)\) is 5.
| Condition for Continuity at \(x=1\) | Value/Expression |
|---|---|
| \(f(1)\) | 5 |
| \(\lim_{x \to 1^-} f(x)\) | \(a + b\) |
| \(\lim_{x \to 1^+} f(x)\) | \(b - a\) |
| Equality for Continuity | \(a + b = b - a = 5\) |
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Continuity at a Point | A function \(f(x)\) is continuous at \(x=c\) if \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)\). | Applied at \(x=1\) to find \(a\) and \(b\). |
| Left-Hand Limit | The limit of \(f(x)\) as \(x\) approaches \(c\) from values less than \(c\) (\(x < c\)). | Calculated using \(f(x) = a+bx\) for \(x < 1\). |
| Right-Hand Limit | The limit of \(f(x)\) as \(x\) approaches \(c\) from values greater than \(c\) (\(x > c\)). | Calculated using \(f(x) = b-ax\) for \(x > 1\). |
| Piecewise Function | A function defined by multiple sub-functions, each applicable to a certain interval of the domain. | The given function is a piecewise function. Continuity must be checked at the boundaries of the intervals. |
Additional Information on Continuous Functions
If a function is continuous over its entire domain, it means there are no breaks, gaps, or jumps in its graph. For polynomial functions like \(a+bx\) and \(b-ax\), they are continuous everywhere within their respective defined intervals (\(x < 1\) and \(x > 1\)). The points where continuity needs careful examination for piecewise functions are the points where the definition of the function changes.
In this problem, since the function is continuous overall, it must be continuous specifically at \(x=1\), which is the only point where its definition switches from one expression to another. The method used above, equating the limits and the function value at the transition point, is the standard approach for ensuring continuity of piecewise functions.
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