Problem 4
Question
Approximate \(\sqrt{98}\) using the appropriate rst derivative to help you. Explain your reasoning.
Step-by-Step Solution
Verified Answer
Using the appropriate first derivative as a helper, the square root of 98 can be approximated as 9.9.
1Step 1: Find Closest Perfect Square to 98 and Its Square Root
A perfect square close to 98 is 100. Its square root is 10.
2Step 2: Describe Square Root Function and Its Derivative
The square root function \(f(x)\) can be represented as \(f(x)=\sqrt{x}\). To find its derivative, \(f'(x)\), you can use the formula: \(f'(x)=1/(2\sqrt{x})\). The derivative of the square root function at the point 100 is \(f'(100)=1/(2*10)=0.05\).
3Step 3: Use the First Derivative to Approximate \(\sqrt{98}\)
The basic formula for linear approximation is: \(f(a)+f'(a)(x-a)\). Here, \(a\) is a point close to \(x\) that we know the value of f of. For this problem, \(a\) is 100 and \(x\) is 98. So, the approximation is \(f(100)+f'(100)(98-100)\). Inserting the values gives \(10+0.05*(98-100) = 10 - 0.05*2 = 10 - 0.1 = 9.9\). Thus, \(\sqrt{98}\) can be estimated as 9.9. Remember, this is only an approximation. The actual value is slightly higher, at approximately 9.899.
Key Concepts
Linear ApproximationFirst DerivativeSquare Root FunctionPerfect Squares
Linear Approximation
Linear approximation is a helpful tool when you need to estimate the value of a function close to a known point. This technique works well for functions that are "smooth," meaning they don't have sharp bends, such as the square root function. The idea is to use the tangent line at a known point to approximate values nearby.
- The formula for a linear approximation is: \[ f(a) + f'(a)(x-a) \]
- Here, \( f(a) \) is the function value at a known point \( a \), \( f'(a) \) is the derivative at \( a \), and \( x \) is the point you wish to approximate.
First Derivative
The first derivative of a function provides critical information about the rate of change of the function at a specific point. In the context of the square root function:
- The derivative, expressed as \( f'(x) = \frac{1}{2\sqrt{x}} \), tells us how steep the function is.
- This slope information is what we use to create the tangent line for a linear approximation.
Square Root Function
The square root function \( f(x) = \sqrt{x} \) is a fundamental mathematical function. Its graph is a smooth curve that slowly increases and is defined for all non-negative numbers.
- One important property is that as \( x \) increases, \( \sqrt{x} \) increases, but at a slowing rate.
- This behavior reflects the idea that larger numbers have their square roots closer together.
Perfect Squares
Perfect squares are numbers that have integers as their square roots, like 1, 4, 9, 16, 25, and so forth. These numbers are very useful in mathematics, especially when approximating other square roots, because they provide reference points.
- For instance, when determining \( \sqrt{98} \), we look to \( \sqrt{100} = 10 \) because 100 is a nearby perfect square.
- This nearby perfect square allows us to use it in the linear approximation formula as our known point \( a \).
Other exercises in this chapter
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