Understanding Algebraic Operations on Continuous Functions
Fundamentals of Adding and Subtracting Continuous Functions
Continuity Preservation in Function Addition
Consider two functions, \(f\) and \(g\), both continuous at a specific point \(a\). The sum of these functions, defined as \((f+g)(x) = f(x) + g(x)\), also maintains continuity at \(a\). This is because the limit of the sum at \(a\) equals the sum of the limits, which matches the function value at \(a\).
Formally, since \(\lim_{x \to a} f(x) = f(a)\) and \(\lim_{x \to a} g(x) = g(a)\), it follows that:
\[ \lim_{x \to a} (f+g)(x) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = f(a) + g(a) = (f+g)(a) \]
This confirms the continuity of the sum at \(x = a\).
Example: Let \(f(x) = x^2\) and \(g(x) = \sqrt{x}\), both continuous at \(x=4\). Show that \(f+g\) is continuous at 4.
Solution: We have \(f(4) = 16\) and \(g(4) = 2\). Since both functions are continuous at 4, their limits equal their function values:
\[ \lim_{x \to 4} f(x) = 16, \quad \lim_{x \to 4} g(x) = 2 \]
Therefore, \[ \lim_{x \to 4} (f+g)(x) = 16 + 2 = 18 = (f+g)(4) \] Hence, \(f+g\) is continuous at \(x=4\).
Continuity in Function Subtraction
Similarly, if \(f\) and \(g\) are continuous at \(a\), their difference \((f-g)(x) = f(x) - g(x)\) is also continuous at \(a\). The limit of the difference equals the difference of the limits, matching the function value at \(a\).
Given \(\lim_{x \to a} f(x) = f(a)\) and \(\lim_{x \to a} g(x) = g(a)\), then:
\[ \lim_{x \to a} (f-g)(x) = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) = f(a) - g(a) = (f-g)(a) \]
This ensures the continuity of the difference at \(x = a\).
Example: Verify that \(f(x) = e^x\) and \(g(x) = \ln(x+2)\) have a continuous difference at \(x=1\).
Solution: Both \(e^x\) and \(\ln(x+2)\) are continuous at \(x=1\). We find:
\[ f(1) = e^1 = e, \quad g(1) = \ln(3) \]
Limits equal function values, so: \[ \lim_{x \to 1} (f-g)(x) = e - \ln(3) = (f-g)(1) \] Thus, the difference is continuous at \(x=1\).
Multiplying and Dividing Continuous Functions
Product of Continuous Functions
When two functions \(f\) and \(g\) are continuous at a point \(a\), their product \((f \cdot g)(x) = f(x) \cdot g(x)\) is also continuous at \(a\). This follows from the property that the limit of a product is the product of the limits.
Since \(\lim_{x \to a} f(x) = f(a)\) and \(\lim_{x \to a} g(x) = g(a)\), it holds that:
\[ \lim_{x \to a} (f \cdot g)(x) = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) = f(a) \cdot g(a) = (f \cdot g)(a) \]
This confirms the product's continuity at \(x = a\).
Example: Show that \(f(x) = \sin x\) and \(g(x) = x^3\) have a continuous product at \(x=0\).
Solution: Both \(\sin x\) and \(x^3\) are continuous everywhere, especially at 0:
\[ f(0) = 0, \quad g(0) = 0 \]
Therefore, \[ \lim_{x \to 0} (f \cdot g)(x) = \lim_{x \to 0} \sin x \cdot \lim_{x \to 0} x^3 = 0 \cdot 0 = 0 = (f \cdot g)(0) \] Hence, the product is continuous at 0.
Continuity in Quotient of Functions
For two functions \(f\) and \(g\) continuous at \(a\), the quotient \(\frac{f}{g}\) is continuous at \(a\) provided \(g(a) \neq 0\). The limit of the quotient equals the quotient of the limits:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} = \frac{f(a)}{g(a)} = \left(\frac{f}{g}\right)(a) \]
This guarantees continuity of the division at \(x = a\) when the denominator is nonzero.
Example: Confirm that \(f(x) = x^2 + 1\) and \(g(x) = x - 1\) have a continuous quotient at \(x=2\).
Solution: Both functions are continuous at 2, and \(g(2) = 1 \neq 0\). We have:
\[ f(2) = 5, \quad g(2) = 1 \]
Thus, \[ \lim_{x \to 2} \frac{f(x)}{g(x)} = \frac{5}{1} = 5 = \left(\frac{f}{g}\right)(2) \] So, the quotient is continuous at \(x=2\).
Continuity in Composite Functions
Preserving Continuity through Composition
Suppose \(f\) and \(g\) are real-valued functions where the composition \((f \circ g)(x) = f(g(x))\) is defined at \(a\). If \(g\) is continuous at \(a\) and \(f\) is continuous at \(g(a)\), then the composite function \(f \circ g\) is continuous at \(a\).
This means the continuity of the inner function at \(a\) and the outer function at the image point \(g(a)\) ensures the continuity of the composition at \(a\).
Example: Prove that \(f(x) = \tan x\) is continuous at \(x = \frac{\pi}{4}\).
Solution: Recall that \(\tan x = \frac{\sin x}{\cos x}\). Both \(\sin x\) and \(\cos x\) are continuous everywhere. Since \(\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \neq 0\), the quotient is defined and continuous at \(\frac{\pi}{4}\). Hence, \(\tan x\) is continuous at \(x = \frac{\pi}{4}\).
Example: Demonstrate that \(f(x) = |\cos x|\) is continuous for all real \(x\).
Solution: Define \(g(x) = \cos x\) and \(h(x) = |x|\). Both \(g\) and \(h\) are continuous everywhere. The function \(f\) can be expressed as the composition \(f = h \circ g\), i.e., \(f(x) = h(g(x)) = |\cos x|\). Since composition of continuous functions is continuous, \(f\) is continuous for all real \(x\).
Quick Reference: Summary of Algebraic Operations on Continuous Functions
Operation | Continuity Condition | Result at Point \(a\) |
|---|---|---|
Addition | \(f\) and \(g\) continuous at \(a\) | \(\lim_{x \to a} (f+g)(x) = f(a) + g(a)\) |
Subtraction | \(f\) and \(g\) continuous at \(a\) | \(\lim_{x \to a} (f-g)(x) = f(a) - g(a)\) |
Multiplication | \(f\) and \(g\) continuous at \(a\) | \(\lim_{x \to a} (f \cdot g)(x) = f(a) \cdot g(a)\) |
Division | \(f\) and \(g\) continuous at \(a\), \(g(a) \neq 0\) | \(\lim_{x \to a} \frac{f}{g}(x) = \frac{f(a)}{g(a)}\) |
Composition | \(g\) continuous at \(a\), \(f\) continuous at \(g(a)\) | \(f \circ g\) continuous at \(a\) |
Glossary of Key Terms
Term | Definition |
|---|---|
Continuous Function | A function \(f\) is continuous at \(a\) if \(\lim_{x \to a} f(x) = f(a)\). |
Limit | The value that a function approaches as the input approaches a point. |
Sum of Functions | The function defined by \((f+g)(x) = f(x) + g(x)\). |
Difference of Functions | The function defined by \((f-g)(x) = f(x) - g(x)\). |
Product of Functions | The function defined by \((f \cdot g)(x) = f(x) \cdot g(x)\). |
Quotient of Functions | The function defined by \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\), where \(g(x) \neq 0\). |
Composite Function | A function formed by applying one function to the result of another, \(f(g(x))\). |
Domain | The set of all input values for which a function is defined. |
Modulus Function | A function that returns the absolute value of its input, denoted \(|x|\). |
Rational Function | A function expressed as the ratio of two polynomials. |
Frequently Asked Questions
What happens when two continuous functions are added or subtracted?
The sum or difference of two functions continuous at a point remains continuous at that point.
Are the product and quotient of continuous functions also continuous?
The product of two continuous functions is continuous. The quotient is continuous if the denominator function is nonzero at the point.
How does composition affect continuity?
If \(g\) is continuous at \(a\) and \(f\) is continuous at \(g(a)\), then the composite function \(f \circ g\) is continuous at \(a\).
Are all rational functions continuous?
Yes, rational functions are continuous wherever their denominator is nonzero.
Can a function defined as an absolute value of a continuous function be continuous?
Yes, since the absolute value function is continuous, composing it with a continuous function results in a continuous function.