Problem 39
Question
Sketch the graph of the function and describe the interval(s) on which the function is continuous. \(f(x)=\frac{x^{2}-16}{x-4}\)
Step-by-Step Solution
Verified Answer
The function is continuous for all real numbers except at \(x=4\). The intervals of continuity are \((-\infty, 4)\) and \( (4, \infty)\).
1Step 1: Simplifying the Function
Simplify the function by factorizing the numerator:\n\[f(x)=\frac{(x-4)(x+4)}{x-4}\]\nThis can be further simplified to:\n\[f(x)=x+4\], whenever \(x\neq4\) because we can't divide by zero.
2Step 2: Plotting the Function
The graph of \(f(x)=x+4\) is a straight line with slope 1 and y-intercept 4. However, remember that our function is not defined at \(x=4\). So the graph of the function is a straight line identical to \(y=x+4\) but with a hole at the point (4, 8).
3Step 3: Determine the Continuity
The function is continuous for all real numbers except at \(x=4\), because at \(x=4\), the function is not defined. This is seen on the graph as a hole at (4, 8). Thus, the intervals of continuity are \((-\infty, 4)\) and \( (4, \infty)\).
Key Concepts
Graphing FunctionsRational FunctionsDiscontinuities in Functions
Graphing Functions
When graphing a function, the main goal is to visualize its behavior across different intervals. For the function provided, \(f(x)=\frac{x^{2}-16}{x-4}\), we can start by simplifying it to \(f(x)=x+4\) for \(xeq4\). At first glance, this appears as a linear function with a slope of 1 and a y-intercept of 4, producing a straight line on the graph. However, it's crucial to remember any restrictions on the domain, like division by zero. In this case, \(x=4\) creates a division by zero, marking a hole in the graph at \((4,8)\). This means that while elsewhere the line \(y=x+4\) is continuous, at \(x=4\) it momentarily breaks.
- Visualize functions accurately by identifying their general shapes and any changes in their behavior.
- Consider simplifications of function expressions for straightforward graph plotting.
- Remember to mark any empty points in the graph, known as holes, where inputs cause undefined behavior.
Rational Functions
Rational functions are expressed as the ratio of two polynomials, such as \(f(x)=\frac{x^{2}-16}{x-4}\). Here, the numerator is \(x^2 - 16\), a polynomial of degree 2, and the denominator is \(x - 4\), a polynomial of degree 1. Simplifying such functions often involves factoring to see if terms can cancel out. This process allowed us to simplify \(f(x)\) to \(x+4\) for all \(xeq 4\). The endeavor with any rational function is to identify points of discontinuity where the denominator equals zero, as these terms cannot be canceled.
- Explore rational functions through factoring and simplification.
- Understand that rational functions are not defined when their denominators equal zero.
- Simplify to the most straightforward form possible while noting any restrictions.
Discontinuities in Functions
Discontinuities occur in functions when there's a break, hole, or undefined point within a graph. The previous example displays what's called a removable discontinuity at \(x=4\), where the function is otherwise well-defined. This happens because we can simplify the function, effectively removing any impact of the discontinuity from analytical calculations, though it still appears graphically. Discontinuities can be due to:
- Vertical asymptotes, occurring when polynomial terms do not cancel out, resulting in infinite limits.
- Removable discontinuities, where factoring allows cancellation with undefined points remaining on the domain level.
- Jump discontinuities, seen in piecewise functions that abruptly switch values.
Other exercises in this chapter
Problem 39
When the price of a glass of lemonade at a lemonade stand was \(\$ 1.75,400\) glasses were sold. When the price was lowered to \(\$ 1.50,500\) glasses were sold
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find \(f^{\prime}(x)\). $$ f(x)=x\left(x^{2}+1\right) $$
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Use the limit definition to find an equation of the tangent line to the graph of \(f\) at the given point. Then verify your results by using a graphing utility
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Find the limit. $$ \lim _{x \rightarrow 1} \frac{\frac{1}{x+4}-\frac{1}{4}}{x} $$
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