Problem 61 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the Intermediate Value Theorem to show that \(\sqrt{x}-\cos x=0\) has a solution between 0 and \(\pi / 2\). Zoom in on the graph of \(y=\sqrt{x}-\cos x\) to find an interval having length \(0.1\) that contains this solution.

Step-by-Step Solution

Answer

There is a solution for \( \sqrt{x} - \cos x = 0 \) between 0.6 and 0.7.

Step 1: Understand the Intermediate Value Theorem (IVT)

The Intermediate Value Theorem states that if a function \( f \) is continuous on a closed interval \([a, b]\) and \( f(a) \) and \( f(b) \) have opposite signs, then there exists at least one \( c \) in the open interval \((a, b)\) such that \( f(c) = 0 \).

Step 2: Define the Function

Define the function \( f(x) = \sqrt{x} - \cos x \). We need to check whether \( f(x) \) is continuous on the interval \([0, \pi/2]\). Both \( \sqrt{x} \) and \( \cos x \) are continuous functions on \([0, \pi/2]\), so \( f(x) \) is continuous as well since it is the difference of two continuous functions.

Step 3: Evaluate the Function at the Endpoints

Calculate \( f(x) \) at the endpoints of the interval. Compute \( f(0) = \sqrt{0} - \cos(0) = 0 - 1 = -1 \). Then compute \( f(\pi/2) = \sqrt{\pi/2} - \cos(\pi/2) = \sqrt{\pi/2} - 0 = \sqrt{\pi/2} \), which is positive. Since \( f(0) < 0 \) and \( f(\pi/2) > 0 \), by the IVT, there is at least one solution \( c \) in \( (0, \pi/2) \) such that \( f(c) = 0 \).

Step 4: Refine the Interval

To find a narrower interval of length \(0.1\), we check the sign of \( f(x) \) at smaller intervals within \((0, \pi/2)\). Test \( f(0.5) = \sqrt{0.5} - \cos(0.5) \). Calculate this numerically or using a calculator: \( \sqrt{0.5} \approx 0.707 \) and \( \cos(0.5) \approx 0.877 \), so \( f(0.5) \approx 0.707 - 0.877 = -0.17 \) (negative).

Step 5: Further Narrow Down the Interval

Next, test \( f(0.6) = \sqrt{0.6} - \cos(0.6) \). Numerically, \( \sqrt{0.6} \approx 0.775 \) and \( \cos(0.6) \approx 0.825 \), so \( f(0.6) \approx 0.775 - 0.825 = -0.05 \) (still negative). Test \( f(0.7) = \sqrt{0.7} - \cos(0.7) \). \( \sqrt{0.7} \approx 0.837 \) and \( \cos(0.7) \approx 0.764 \), so \( f(0.7) \approx 0.837 - 0.764 = 0.073 \) (positive). This indicates a root between 0.6 and 0.7.

Key Concepts

Continuous function

Imagine a smooth road that does not break or stop anywhere—it just flows nicely all along. A continuous function is kind of like that road. In math, a continuous function does not have any holes, jumps, or breaks in its graph.

For example, the function \( \sqrt{x} \) is continuous everywhere it is defined (meaning we don't take the square root of a negative number in this case). The cosine function \( \cos x \) is also continuous because it smoothly oscillates between -1 and 1. When you combine these two functions by subtraction, as in \( f(x) = \sqrt{x} - \cos x \), the new function remains continuous.

• The result is continuous because summing or subtracting continuous functions still keeps the outcome continuous.
• Because both \( \sqrt{x} \) and \( \cos x \) work perfectly on the interval \([0, \pi/2]\), their difference is also continuous over this whole range.

A continuous function is like an unbroken thread stretching across an interval, which is key to applying the Intermediate Value Theorem.

Root finding

Finding a root essentially means locating where the function hits the horizontal axis or, in simpler terms, where the function equals zero.

This involves setting up the function \( f(x) \) you believe (or have calculated) has some points where \( f(x) = 0 \). In our example, we are working with \( f(x) = \sqrt{x} - \cos x \). To apply the Intermediate Value Theorem (IVT), we need two values, \( a \) and \( b \), where \( f(a) \) and \( f(b) \) have different signs. This means that one is positive, and the other is negative.

• First, check at the start point, \( f(0) = \sqrt{0} - \cos(0) = -1 \), a negative number.
• Then, at the endpoint \( \pi/2 \), \( f(\pi/2)=\sqrt{\pi/2} - \cos(\pi/2) \approx \sqrt{\pi/2} \), a positive number.

With opposite signs at these points, the Intermediate Value Theorem assures us there’s an \( x \) between 0 and \( \pi/2 \) that makes \( f(x) = 0 \). This is because the line must cross the horizontal axis to move from negative to positive.

Interval narrowing

Now comes the exciting process of closing in on the root—much like a game of hot and cold. We know there is a solution on \( (0, \pi/2) \) to \( f(x) = 0 \). But let's get more precise.

The task is to narrow it into an interval no longer than 0.1. Think of it like gradually zooming in with a magnifying glass. We start testing smaller intervals within \( (0, \pi/2) \). Calculate the function at various points and note whether they are positive or negative.

• Try \( f(0.5) \approx -0.17 \), which is negative.
• Next, \( f(0.6) \approx -0.05 \), still negative.
• Then \( f(0.7) \approx 0.073 \), a positive number.

By observing this change, we know the root lies between 0.6 and 0.7. The sign switch between negative and positive pinpoints the interval where zero must be hiding, making this our narrowed space to find the root accurately. This technique is a focused search, using the science of continuity and sign-changes, to effectively isolate solutions.