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Chapter 7: Problem 10

Suppose \(f\) and \(g\) are continuous on \([a, b]\) and that \(f(a)\) \(g(b) .\) Prove that \(f(x)=g(x)\) for some \(x\) in \([a, b] .\) (If your proof isn't very short, it's not the right one.)

Short Answer

Expert verified
By the Intermediate Value Theorem, \(f(x) = g(x)\) for some \(x\) in \([a, b]\).

Step by step solution

01

- Understand the Given Information

The functions \(f\) and \(g\) are continuous on the interval \([a, b]\), and you are given that \(f(a) < g(a)\) and \(f(b) > g(b)\).
02

- Define a New Function

Define a new function \(h(x) = f(x) - g(x)\). Since \(f\) and \(g\) are continuous, \(h\) will also be continuous on \([a, b]\).
03

- Evaluate the New Function at the Endpoints

Evaluate \(h\) at \(a\) and \(b\): \(h(a) = f(a) - g(a) < 0\) since \(f(a) < g(a)\). \(h(b) = f(b) - g(b) > 0\) since \(f(b) > g(b)\).
04

- Apply the Intermediate Value Theorem

By the Intermediate Value Theorem, since \(h(x)\) is continuous on \([a, b]\) and changes sign (from negative to positive), there must be some point \(c\) in \((a, b)\) where \(h(c) = 0\). In other words, \(f(c) - g(c) = 0\), or \(f(c) = g(c)\).
05

Conclusion

Therefore, there exists some \(x = c\) in \([a, b]\) such that \(f(x) = g(x)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity
The concept of continuity is crucial when dealing with functions in calculus. Simply put, a function is continuous over an interval if there are no breaks, jumps, or holes in its graph.
For the given problem, we know that both functions, \(f\) and \(g\), are continuous over the closed interval \[a, b\]. Since there are no disruptions in their graphs within this interval, we can apply powerful mathematical tools like the Intermediate Value Theorem.
Let's begin by understanding that continuity means the values of \(f(x)\) and \(g(x)\) smoothly transition from one point to the next within the given interval. This property allows us to define a new function \(h(x) = f(x) - g(x)\) and know with confidence that \(h(x)\) will also be continuous on \[a, b\].
Function Difference
A critical step in solving the problem is defining a new function based on the difference between \(f\) and \(g\).
By defining \(h(x) = f(x) - g(x)\), we transform the original problem into a simpler one. Since both \(f\) and \(g\) are continuous, the new function \(h\) inherits this continuity.
Evaluating \(h\) at the endpoints gives us:
  • \(h(a) = f(a) - g(a)\), which is less than 0 (because \(f(a) < g(a)\)).
  • \(h(b) = f(b) - g(b)\), which is greater than 0 (because \(f(b) > g(b)\)).
This setup shows that \(h(x)\) changes from negative at \(a\) to positive at \(b\), setting the stage for the next key point.
Endpoint Evaluation
Now, let's focus on the evaluations at the endpoints of the interval and the application of the Intermediate Value Theorem.
We evaluated \(h(a)\) and \(h(b)\) as part of the previous steps. These evaluations showed that \(h(a) < 0\) and \(h(b) > 0\). Because \(h(x)\) is continuous over \[a, b\], it means the function must cross the x-axis somewhere within the interval \((a, b)\).
This is a perfect situation to use the Intermediate Value Theorem, which states: If a function is continuous on a closed interval and takes different signs at the endpoints, there must be at least one point within the interval where the function equals zero.
In our context: Since \(h(x)\) is continuous and there are sign changes from \(a\) to \(b\), there exists some point \(c\) in \((a, b)\) where \(h(c) = 0\). This means \(f(c) - g(c) = 0\), leading us to conclude that \(f(c) = g(c)\).
Therefore, we've successfully proven that there exists some point \(x = c\) in \[a, b\] where \(f(x) = g(x)\).

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Most popular questions from this chapter

Suppose that \(f \text { is a contimous function on } |-1.1]\) such that \(x^{2}+(f(x))^{2}=1\) for all \(x\). (This means that \((x, f(x)\) ) always lies on the unit circle.) Show that either \(f(x)=\sqrt{1-x^{2}}\) for all \(x,\) or else \(f(x)=-\sqrt{1-x^{2}}\) for all \(x\).

(a) Suppose that \(f\) is continuous on \((a, b)\) and \(\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow b^{-}} f(x)=\infty\). Prowe that \(f\) has a minimum on all of \((a, b)\) (b) Prowe the corresponding result when \(a=-\infty\) and/or \(b=\infty\).

(a) Prove that there does not exist a continuous function \(f\) defined on \(\mathbf{R}\) which takes on every value exactly twice. Hint: II \(f(a)=f(b)\) for \(af(a)\) for all \(x\) in \((a, b)\) or \(f(x)b\) (b) Refine part (a) by proving that there is no continuous function \(f\) which takes on each value either 0 times or 2 times, i.e., which takes on exactly twice each value that it does tiake on. Hint: The previous lint implies that \(f\) las cither a natxinum or a niminum value (which must be taken on twice). What can be said about values close to the maximum value? (c) Find a continuous function \(f\) which takes on every value exactly 3 times. More generally, find one which tikes on every valur exactly \(n\) times, if \(n\) is odd. (d) Prove that if \(n\) is even, then there is no continuous \(f\) which takes on every value exactly \(n\) times. Hint: To treat the case \(n=4\), for example, let \(f\left(x_{1}\right)=f\left(x_{2}\right)=f\left(x_{3}\right)=f\left(x_{4}\right) .\) Ifuen either \(f(x)>0\) for all \(x\) in two of the three intervals \(\left(x_{1}, x_{2}\right),\left(x_{2}, x_{3}\right),\left(x_{3}, x_{4}\right),\) or che \(f(x)<0\) for all \(x\) in two of these three intervals.

Suppose that \(f\) and \(g\) are continuous, that \(f^{2}=g^{2},\) and that \(f(x) \neq 0\) for all \(x\). Prove that either \(f(x)=g(x)\) for all \(x\), or else \(f(x)=-g(x)\) for all \(x\).

(a) Suppose that \(f\) is continuous, that \(f(x)=0\) only for \(x=a,\) and that \(f(x)>0\) for some \(x>a\) as well as for some \(x0\) for some \(x>a\) and \(f(x)<0\) for some \(x

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