Problem 4
Question
Let $$ f(x)=\sin x-x, \quad-1 \leq x \leq 1 $$ (a) Graph \(y=f(x)\) for \(-1 \leq x \leq 1\). (b) Use the intermediate-value theorem to conclude that $$ \sin x=x $$ has a solution in \((-1,1)\).
Step-by-Step Solution
Verified Answer
The equation \( \sin x = x \) has at least one solution in the interval \((-1, 1)\) by the Intermediate-Value Theorem.
1Step 1: Understand the Function
We are given the function \( f(x) = \sin x - x \) over the interval \( -1 \leq x \leq 1 \). This function is the difference between the sine function and the identity function, which will help in determining the behavior of the graph and analyzing the roots.
2Step 2: Graph the Function
To graph \( y = f(x) = \sin x - x \) over \( -1 \leq x \leq 1 \), calculate key points and sketch. For example, at \( x = 0 \), \( f(x) = \sin(0) - 0 = 0 \). Plot additional points like \( x = -1 \) and \( x = 1 \) to observe the changes. Sketch the curve considering the sinusoidal nature of \( \sin x \) and the linear nature of \( -x \).
3Step 3: Apply the Intermediate-Value Theorem
The Intermediate-Value Theorem states that if a function \( g \) is continuous over an interval \([a, b]\) and \( g(a) \) and \( g(b) \) have opposite signs, then there exists a \( c \) in \((a, b)\) such that \( g(c) = 0 \). Observe \( f(-1) = \sin(-1) + 1 \) and \( f(1) = \sin(1) - 1 \). Since \( \sin(-1) + 1 > 0 \) and \( \sin(1) - 1 < 0 \), \( f(x) \) changes sign, confirming that a root exists in \((-1, 1)\).
4Step 4: Conclusion
Because \( f(x) = \sin x - x \) is continuous on \([-1, 1]\) and changes sign from positive to negative, the Intermediate-Value Theorem guarantees a solution to \( \sin x = x \) within the interval \((-1, 1)\).
Key Concepts
continuous functionsgraphing functionsroots of equations
continuous functions
Understanding continuous functions can be crucial when dealing with problems involving the Intermediate-Value Theorem. A function is continuous if there are no breaks, jumps, or gaps in its graph over a given interval. This means you can draw the graph of the function from start to finish without lifting your pencil from the paper. For the function \( f(x) = \sin x - x \) over the interval \([-1, 1]\), continuity is important to verify before applying certain theorems.
The sine function, \( \sin x \), is well-known to be continuous everywhere, while the linear function \(-x\) is also continuous. Therefore, the difference \( \sin x - x \) remains continuous over any real number, especially within the bounded interval \([-1, 1]\).
When dealing with continuous functions, it is crucial to ensure their continuity over the entire interval of interest. This characteristic allows us to apply the Intermediate-Value Theorem effectively, conclusively showing that certain equations have solutions within specific intervals.
The sine function, \( \sin x \), is well-known to be continuous everywhere, while the linear function \(-x\) is also continuous. Therefore, the difference \( \sin x - x \) remains continuous over any real number, especially within the bounded interval \([-1, 1]\).
When dealing with continuous functions, it is crucial to ensure their continuity over the entire interval of interest. This characteristic allows us to apply the Intermediate-Value Theorem effectively, conclusively showing that certain equations have solutions within specific intervals.
graphing functions
Graphing functions can give us a visual representation to better understand the behavior of the function within a certain domain. When graphing the function \( y = f(x) = \sin x - x \) over the interval \(-1 \leq x \leq 1\), it's important to consider both component functions: the sine function and the linear x function.
The sine function oscillates between -1 and 1, generating a wave-like curve. The linear function \(-x\) creates a diagonal line, sloping downward from left to right. By plotting key points, such as where \( x = -1, 0, \) and \( 1 \), we can sketch the resulting curve of \( f(x) \).
At \( x = 0 \), \( f(x) = \sin(0) - 0 = 0 \), so the graph passes through the origin. As x moves toward 1 or -1, the function's value will increase or decrease based on the effect of both the sine and linear functions.
You’ll see that the resulting graph of \( \sin x - x \) smoothly blends the oscillating nature of sine with the straight decline of \(-x\), making it continuous and smooth.
The sine function oscillates between -1 and 1, generating a wave-like curve. The linear function \(-x\) creates a diagonal line, sloping downward from left to right. By plotting key points, such as where \( x = -1, 0, \) and \( 1 \), we can sketch the resulting curve of \( f(x) \).
At \( x = 0 \), \( f(x) = \sin(0) - 0 = 0 \), so the graph passes through the origin. As x moves toward 1 or -1, the function's value will increase or decrease based on the effect of both the sine and linear functions.
You’ll see that the resulting graph of \( \sin x - x \) smoothly blends the oscillating nature of sine with the straight decline of \(-x\), making it continuous and smooth.
roots of equations
Finding the roots of equations involves determining where a particular function equals zero. This process is essential when trying to solve equations like \( \sin x = x \). To find the roots using the function \( f(x) = \sin x - x \), we set \( f(x) = 0 \) and look for values of x that satisfy this equation.
By graphing \( f(x) \) and observing where it crosses the x-axis, we pinpoint potential roots, where the function changes from positive to negative. This change implies a zero because the graph transitions from above the x-axis to below it (or vice versa).
More formally, the Intermediate-Value Theorem aids this search. If \( f(x) \) is continuous on the interval \([-1, 1]\), and we observe \( f(-1) > 0 \) and \( f(1) < 0 \), a root, or zero, must exist between \(-1\) and \(1\). The function \( \sin x - x \) indeed satisfies these conditions, meaning \( \sin x = x \) has at least one solution within the interval. This provides confidence in the existence of roots and thus the solutions to our original equation.
By graphing \( f(x) \) and observing where it crosses the x-axis, we pinpoint potential roots, where the function changes from positive to negative. This change implies a zero because the graph transitions from above the x-axis to below it (or vice versa).
More formally, the Intermediate-Value Theorem aids this search. If \( f(x) \) is continuous on the interval \([-1, 1]\), and we observe \( f(-1) > 0 \) and \( f(1) < 0 \), a root, or zero, must exist between \(-1\) and \(1\). The function \( \sin x - x \) indeed satisfies these conditions, meaning \( \sin x = x \) has at least one solution within the interval. This provides confidence in the existence of roots and thus the solutions to our original equation.
Other exercises in this chapter
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