Problem 10
Question
Suppose that \(f^{\prime \prime}\) exists and is continuous on \([1,2] .\) Suppose also that \(\left|f^{\prime \prime}(x)\right| \leq \frac{1}{4}\) for all \(x\) in \((1,2) .\) Prove that if we use the linearization \(y=L(x)\) of \(y=f(x)\) at \(x=1\) as an approximation of \(y=f(x)\) near \(x=1,\) then our estimated value of \(f(1.2)\) is guaranteed to have an accuracy of at least \(0.01,\) i.e., our estimate will lie within 0.01 units of the true value.
Step-by-Step Solution
Verified Answer
The error of the linear approximation at \( x = 1.2 \) is at most 0.005, ensuring an accuracy within 0.01 of the true value.
1Step 1: Understand Taylor's Theorem
Taylor's theorem states that for a function \( f \) that is sufficiently smooth at a point \( a \), the function can be approximated as \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(c)(x-a)^2}{2} \) for some \( c \) between \( a \) and \( x \). Here, we are using a linear approximation \( L(x) = f(a) + f'(a)(x-a) \) at \( x = 1 \), so the error term is \( \frac{f''(c)(x-a)^2}{2} \). The remainder term is of the order of \( (x-a)^2 \).
2Step 2: Apply Bounds on the Second Derivative
We are given that \( |f''(x)| \leq \frac{1}{4} \) for all \( x \) in \((1,2)\). This indicates that the worst-case error when approximating \( f(x) \) using its linearization at \( x = 1 \) can be bounded by \( \left| \frac{f''(c)(x-1)^2}{2} \right| \leq \frac{1}{4} \times (x-1)^2\).
3Step 3: Calculate the Error at x = 1.2
We are approximating \( f(1.2)\) using \( L(x)\). The error is \( \left| \frac{f''(c)(1.2-1)^2}{2} \right| \). Substitute the values to get \( \left| \frac{f''(c) \times (0.2)^2}{2} \right| \leq \frac{1}{4} \times \frac{0.04}{2} = 0.005\).
4Step 4: Confirm the Error Falls Within Acceptable Range
The maximum error calculated is \( 0.005 \), which is less than \( 0.01 \). Therefore, the linear approximation \( L(x) \) for \( f(x) \) at \( x = 1.2 \) is within \( 0.01 \) units of the true value, meeting the required accuracy condition.
Key Concepts
LinearizationRemainder TermSecond Derivative BoundsError Estimation
Linearization
Linearization is a concept used to simplify the analysis of functions by approximating them with linear functions. It's like zooming in and observing a curve and noticing that it appears straight. The essence of linearization lies in representing a function close to a certain point as a straight line. This is a powerful tool in calculus, especially when dealing with complex functions.
- The linearization of a function \( f \) at a point \( a \) is given by the formula \( L(x) = f(a) + f'(a)(x-a) \).
- This linear function is an approximation of \( f(x) \) that's easiest to compute and understand near \( x = a \).
Remainder Term
The Remainder Term is a crucial part of Taylor's Theorem. It provides a measure of how precise our linear approximation of a function is compared to the actual function value. Think of it as the error gap between the approximation and the reality.
- For a linear approximation, the remainder term is formulated as \( \frac{f''(c)(x-a)^2}{2} \).
- Here, \( c \) is a point between \( a \) and \( x \) that ensures this term accounts for all potential discrepancies.
Second Derivative Bounds
Bounding the second derivative is fundamental to controlling the error in our linear approximations. It is much like setting boundaries within which our function behaves predictably. When using Taylor's Theorem, we aim to understand how quickly the function's change is changing, and we use this to our advantage.
- The bound says \( |f''(x)| \leq \frac{1}{4} \) for \( x \) between 1 and 2 in our example scenario.
- Having this bound provides a limit to the curvature (or bendiness) of the function.
Error Estimation
Error estimation is the art of betting how far off our approximation might be from the actual outcome. In Taylor's theorem applications, this is calculating how much the linear version of a function differs from its real form.
- Error estimation involves evaluating \( \left| \frac{f''(c)(x-a)^2}{2} \right| \).
- We substitute values to give us a concrete estimate, like \( 0.005 \) in our exercise.
Other exercises in this chapter
Problem 10
Find all critical points and identify them as local maximum points, local minimum points, or neither. $$ f(x)=\left|x^{2}-121\right| $$
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Compute the following limits. $$ \lim _{t \rightarrow 0^{+}}\left(\frac{1}{t}+\frac{1}{\sqrt{t}}\right)(\sqrt{t+1}-1) $$
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Find \(f(x)\) if \(f^{\prime}(x)=e^{-x}\) and \(f(0)=2\).
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For any real number x there is a unique integer n such that \(n \leq x
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