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Magazine Chapter 6: Problem 23
Rolle's theorem is applicable for the function \(f(x)=(x-1)|x|+|x-1|\) in the interval (a) \([0,1]\) (b) \(\left[\frac{1}{4}, \frac{3}{4}\right]\) (c) \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) (d) \(\left[\frac{1}{5}, \frac{6}{7}\right]\)
Short Answer
Expert verified
Rolle's theorem is applicable only for the interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
Step by step solution
01
Analyze the function
Firstly, observe that this function is continuous at \(x=0\), but if \(x\geq0\), \(f(x)=x^2-1\) and if \(x<0\), \(f(x)=-x^2+1\). So, the function is differentiable everywhere, except at \(x=1\). To apply Rolle's theorem, it should be checked whether \(f(a) = f(b)\) for every given interval.
02
Check the interval [0,1]
For the interval [0,1], verify if \(f(0) = f(1)\). Substituting the values into the function gives \(f(0) = (-1)|0| + |0-1| = 0+1 = 1\) and \(f(1) = (1-1)|1| + |1-1| = 0+0 = 0\). So, \(f(0) \neq f(1)\). Therefore, Rolle's theorem cannot be applied for this interval.
03
Check the interval \(\left[\frac{1}{4}, \frac{3}{4}\right]\)
For the interval \(\left[\frac{1}{4}, \frac{3}{4}\right]\), observe that \(f\left(\frac{1}{4}\right) = \left(\frac{1}{4}-1\right)\left|\frac{1}{4}\right| + \left|\frac{1}{4}-1\right| = -\frac{3}{4}|\frac{1}{4}| + \frac{3}{4} = -\frac{3}{16} + \frac{3}{4} = \frac{9}{16}\) and \(f\left(\frac{3}{4}\right) = \left(\frac{3}{4}-1\right)\left|\frac{3}{4}\right| + \left|\frac{3}{4}-1\right| = -\frac{1}{4}|\frac{3}{4}| + \frac{1}{4} = -\frac{3}{16} + \frac{1}{4} = \frac{1}{16}\). Since \(f\left(\frac{1}{4}\right) \neq f\left(\frac{3}{4}\right)\), Rolle's theorem is not applicable for this interval.
04
Check the interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)
For the interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\), \(f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}-1\right)\left|-\frac{1}{2}\right| + |-(-\frac{1}{2}-1)| = -\frac{3}{2}*\frac{1}{2} + 1.5 = -\frac{3}{4} + 1.5 = 0.75\) and \(f\left(\frac{1}{2}\right) = \left(\frac{1}{2}-1\right)*\frac{1}{2} + 1.5 = -\frac{1}{4}*\frac{1}{2} + 1.5 = 0.75\). Since \(f\left(-\frac{1}{2}\right) = f\left(\frac{1}{2}\right)\), Rolle's theorem can be applied for this interval.
05
Check the interval \(\left[\frac{1}{5}, \frac{6}{7}\right]\)
For the interval \(\left[\frac{1}{5}, \frac{6}{7}\right]\), \(f\left(\frac{1}{5}\right) = (\frac{1}{5}-1)\left|\frac{1}{5}\right| + \left|\frac{1}{5}-1\right| = -\frac{4}{5}*\frac{1}{5} + 0.8 = -\frac{4}{25} + 0.8 = 0.64\) and \(f\left(\frac{6}{7}\right) = (\frac{6}{7}-1)\left|\frac{6}{7}\right| + \left|\frac{6}{7}-1\right| = -\frac{1}{7}*\frac{6}{7} + 0.143 = -\frac{6}{49} + 0.143 = 0.0216\). Since \(f\left(\frac{1}{5}\right) \neq f\left(\frac{6}{7}\right)\), Rolle's theorem cannot be applied for this interval.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a fundamental concept in calculus that calculates the rate at which a function is changing at any given point. It's like measuring the slope of a curve at a specific spot.
To understand differentiation, think of a function as a smooth curve. The derivative of a function at a point tells us how steep the curve is at that point.
Using calculus, we find that the derivative of a function reveals important characteristics about its behavior:
To understand differentiation, think of a function as a smooth curve. The derivative of a function at a point tells us how steep the curve is at that point.
Using calculus, we find that the derivative of a function reveals important characteristics about its behavior:
- Critical Points: These are where a function might change from increasing to decreasing, or vice versa.
- Slope: The derivative gives us the slope of the tangent line to the curve at any point.
- Rate of Change: Differentiation helps us understand how functions increase or decrease in real-world scenarios.
Continuous Functions
A continuous function is one where small changes in the input result in small changes in the output. In simple terms, you can draw a continuous function without lifting your pencil off the paper.
Continuity is essential for many mathematical theorems, including Rolle's Theorem. For a function to be applicable under Rolle's Theorem, it must be continuous on the given closed interval.
Continuity is essential for many mathematical theorems, including Rolle's Theorem. For a function to be applicable under Rolle's Theorem, it must be continuous on the given closed interval.
- Definition: A function \(f(x)\) is continuous at a point \(a\) if \(\lim_{{x \to a}} f(x) = f(a)\).
- Importance: Continuous functions are predictable and smooth, making them easier to analyze.
- Applications: Continuity is crucial in physics, economics, and engineering because it models real-world phenomena accurately.
Interval Analysis
Interval analysis in calculus involves examining a function's behavior over a specific range, or interval, on the x-axis. This analysis is vital when applying Rolle's Theorem, as it asks us to ensure certain properties hold over the interval.
Rolle's Theorem specifically requires:
Rolle's Theorem specifically requires:
- The function must be continuous over the closed interval \([a, b]\).
- The function must be differentiable over the open interval \((a, b)\).
- The function's values at the endpoints must be equal, meaning \(f(a) = f(b)\).
