Problem 65 - University Calculus: Early Transcendentals

Question

Show that \(|\cos x-1| \leq|x|\) for all \(x\) -values. (Hint: Consider \(f(t)=\cos t \text { on }[0, x] .)\).

Step-by-Step Solution

Answer

The inequality holds due to the Mean Value Theorem and the derivative bounds of the cosine function.

Step 1: Understand the Function

We are given the function \( f(t) = \cos t \) and need to prove that \( |\cos x - 1| \leq |x| \) for all \( x \). The inequality needs to be true for both positive and negative values of \( x \), but for simplicity, consider \( x \geq 0 \). If \( x < 0 \), the argument is analogous due to the even nature of cosine.

Step 2: Analyze at Zero

At \( t=0 \), \( \cos 0 = 1 \). Therefore, \( f(0) = 1 \). \( |\cos x - 1| \) becomes \( |\cos x - \cos 0| \), which can be associated with the property of derivatives and Lipschitz continuity where functions like cosine do not exceed their input value in change rate.

Step 3: Apply Mean Value Theorem

Use the Mean Value Theorem, which states that for a function \( f \) continuous on \([a, b]\) and differentiable on \((a, b)\), there is at least one point \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b-a} \). Apply this between 0 and \( x \):\[ \cos x - 1 = \cos \xi \cdot (x - 0) \] for some \( \xi \in (0, x) \).

Step 4: Evaluate the Derivative

The derivative of \( f(t) = \cos t \) is \( f'(t) = -\sin t \). So, the absolute value of the derivative \( |-\sin \xi| \leq 1 \) because the sine function ranges between -1 and 1. Hence, \( |\cos x - 1| = |-\sin \xi||x| \leq |x| \) for some \( \xi \in (0, x) \).

Step 5: Final Argument and Verification

Since for any \( x \geq 0 \), it is true that \( |\cos x - 1| \leq \sin \xi \cdot x \leq x \), and because the analysis is the same for negative \( x \) due to symmetry of \( \cos(t) \), we conclude that \( |\cos x - 1| \leq |x| \) for all \( x \).

Key Concepts

Mean Value Theorem

The Mean Value Theorem (MVT) is an incredibly useful tool in calculus. It's like a bridge connecting the average change of a function to its instantaneous change at some point within the interval. Here’s how it works:

• For a function to apply this theorem, it must be continuous on the closed interval \[a, b\] and differentiable on the open interval \(a, b\).
• MVT guarantees a point \(c\) in \(a, b\) where the derivative at \(c\), \(/f'(c)/\), equals the average rate of change over \[a, b\].

In simpler words, there exists at least one point in the interval where the tangent (derivative) is perfectly parallel to the secant line connecting the endpoints. Consider a smooth hill road; there'll be a spot where your incline matches the total ascent rate over the hill. In terms of the exercise about cosine, the theorem helps in bounding changes of \/\cos\ using sine since \/\sin\xi\ indicates instantaneous rates of change over a range.
This principle is especially handy when analyzing functions like trigonometric ones over intervals to find maximum or minimum values or to bound them.

Trigonometric Functions

Trigonometric functions are fundamental in connecting angles to ratio aspects of a triangle. The primary ones include sine, cosine, and tangent, each having unique properties:

• Cosine, \(/\cos x/\), relates the adjacent side to the hypotenuse in a right triangle.
• It has an even function property, therefore, \( \cos(-x) = \cos x \) and periodic behavior repeating every \(2\pi\).

Within the problem context, \/\cos x\ is examined for its behavior and relationship to inputs like \(x.\) It's key to understand that as \/\x\ proceeds towards zero, \/\cos x\ approaches 1. Hence, any deviation \/\|\cos x - 1|\ gives insight into the small oscillations against this background.
This "deviation" acts as an exploration of cosine's continuous, yet bounded behavior and supports inequality explorations, where functions barely deviate from constants over small ranges.

Derivative Applications

Derivatives measure instantaneous rates of change. This makes them crucial for understanding the way functions behave globally from local behavior insights.

• In trigonometric contexts, consider how \/\f(t) = \cos t\ leads to a derivative of \/\f'(t) = -\sin t.\
• Notice how the maximum absolute value of \/\sin t\ is 1. This makes the derivative very easy to work with in inequalities, as it sets limits like \/\|\cos x - 1| \leq |x|.\

The derivative can thus gauge maximum potential differences the logarithmic or sinusoidal function might cause. This helps determine if function changes align with external values like inputs or intervals, firmly grounding these mathematical constructs in analysis.
Remember, the point of derivative-based inequalities is that they allow us to make broad generalizations about a function's limits, illustrated perfectly by the inequality shown between \/\cos x\ and \(/x/\).