Ch. 2 - Limits and Continuity

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 2 - Limits and Continuity

Problema 2.5.52

Capitolo 2, Problema 2.5.52

Explain why the equation cos x = x has at least one solution.

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Consider the function \( f(x) = \cos x - x \). We want to show that this function has at least one root, meaning there is at least one value of \( x \) for which \( f(x) = 0 \).

To apply the Intermediate Value Theorem, we need to find two values, \( a \) and \( b \), such that \( f(a) \) and \( f(b) \) have opposite signs. This will indicate that there is at least one root in the interval \([a, b]\).

Evaluate \( f(x) \) at \( x = 0 \): \( f(0) = \cos(0) - 0 = 1 \). So, \( f(0) = 1 \).

Evaluate \( f(x) \) at \( x = \pi/2 \): \( f(\pi/2) = \cos(\pi/2) - \pi/2 = 0 - \pi/2 = -\pi/2 \). So, \( f(\pi/2) < 0 \).

Since \( f(0) > 0 \) and \( f(\pi/2) < 0 \), by the Intermediate Value Theorem, there must be at least one value \( c \) in the interval \((0, \pi/2)\) such that \( f(c) = 0 \). Therefore, the equation \( \cos x = x \) has at least one solution.

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Intermediate Value Theorem

The Intermediate Value Theorem states that if a continuous function takes on two values at two points, it must also take on every value between those two points. This theorem is crucial for proving the existence of solutions to equations like cos x = x, as it guarantees that if the function changes sign over an interval, there is at least one root in that interval.

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Fundamental Theorem of Calculus Part 1

Continuity of Functions

A function is continuous if there are no breaks, jumps, or holes in its graph. The functions cos x and f(x) = x are both continuous over the real numbers. This property is essential for applying the Intermediate Value Theorem, as it ensures that the function behaves predictably and that solutions can be found within specified intervals.

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Intro to Continuity

Behavior of the Functions

To analyze the equation cos x = x, it's important to understand the behavior of the functions involved. The cosine function oscillates between -1 and 1, while the line y = x increases without bound. By examining the intersection points of these two graphs, we can conclude that there must be at least one point where they are equal, confirming the existence of a solution.

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Graphs of Exponential Functions

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Slope of a Curve at a Point

In Exercises 7–18, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P.

y=x³−3x²+4, P(2,0)

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Using limθ→0 sin θ / θ = 1

Find the limits in Exercises 23–46.

limx→0 (cos²x − cos x) / x²

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Limits of quotients

Find the limits in Exercises 23–42.

limx→−5 (x² + 3x − 10) / x + 5

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Suppose that a function f(x) is defined for all real values of x except x=c. Can anything be said about the existence of limx→c f(x)? Give reasons for your answer.

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Limits of Rational Functions

In Exercises 13–22, find the limit of each rational function (a) as x → ∞ and (b) as x → −∞. Write ∞ or −∞ where appropriate.

f(x) = (2x + 3)/(5x + 7)

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Removable discontinuity Give an example of a function f (x) that is continuous for all values of x except x = 2, where it has a removable discontinuity. Explain how you know that f is discontinuous at x = 2, and how you know the discontinuity is removable.

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