Chapter 11: Problem 48
Let \(f\) be a twice differentiable function and fix a value of \(x\) (a) Show that $$\lim _{\Delta \rightarrow 0} \frac{f(x+h)-f(x-h)}{2 h}=f^{\prime}(x)$$
Short Answer
Expert verified
We rewrote the given expression as a limit with h approaching 0, broke up the expression into two separate fractions, and introduced a substitution \(u = -h\). By applying the limit laws and recognizing the definition of the derivative, we simplified the expression to show that \(\lim_{\Delta \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h} = f'(x)\), completing the proof.
Step by step solution
01
Rewrite the expression
Rewrite the given expression to show that Δ → 0 is the same as h → 0:
\[
\lim_{\Delta \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h} = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}
\]
02
Use the definition of the derivative
Recall the limit definition of the derivative of a function f at a point x:
\[
f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}
\]
Our goal is to show that the given expression is equal to f'(x) by manipulating the expression in such a way that it matches the definition of the derivative.
03
Manipulate the expression
Break up the given expression into two separate fractions:
\[
\lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h} = \lim_{h \rightarrow 0} \frac{1}{2}\left(\frac{f(x+h) - f(x)}{h}\right) + \frac{1}{2}\left(\frac{f(x) - f(x-h)}{h}\right)
\]
Note that the first term contains the definition of the derivative of f at x, while the second term is similar but not the same.
04
Introduce a substitution
Let's consider the substitution \(u = -h\). Then, as \(h \rightarrow 0\), we also have \(u \rightarrow 0\). Now, rewrite the second term in the expression using this substitution:
\[
\lim_{u \rightarrow 0} \frac{f(x) - f(x+u)}{u}
\]
05
Apply the limit laws
Using the limit laws, we can take the limit of both terms separately:
\[
\lim_{h \rightarrow 0} \frac{1}{2}\left(\frac{f(x+h) - f(x)}{h}\right) + \lim_{u \rightarrow 0} \frac{1}{2}\left(\frac{f(x) - f(x+u)}{-u}\right)
\]
06
Recognize the definition of the derivative
Observe that both terms in the above expression contain the definition of the derivative of f at x. Thus, we can replace them by f'(x):
\[
\frac{1}{2}f'(x) + \frac{1}{2}(-f'(x))
\]
07
Simplify the expression
Now, simplify the expression to obtain the desired result:
\[
\frac{1}{2}f'(x) - \frac{1}{2}f'(x) = f'(x)
\]
We have shown that
\[
\lim_{\Delta \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h} = f'(x)
\]
which completes the proof.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
In calculus, the derivative represents the rate at which a function's value changes as its input changes. This concept is fundamental, as it provides a way to understand how a function behaves locally—that is, at or near a specific point on its graph.
### Calculating the DerivativeThe derivative of a function \( f \) at a point \( x \) is expressed as \( f'(x) \). It is found using the limit definition:
### ExampleTo solidify this understanding, apply the derivative definition to the function \( f(x) = x^2 \):
Understanding derivatives is crucial for analyzing and predicting patterns of change in various fields, such as physics and economics.
### Calculating the DerivativeThe derivative of a function \( f \) at a point \( x \) is expressed as \( f'(x) \). It is found using the limit definition:
- \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)
### ExampleTo solidify this understanding, apply the derivative definition to the function \( f(x) = x^2 \):
- \( f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \)
- This simplifies to \( 2x \) when \( h = 0 \).
Understanding derivatives is crucial for analyzing and predicting patterns of change in various fields, such as physics and economics.
Limit
A limit in calculus is used to define the value that a function \( f(x) \) approaches as the input \( x \) approaches some value. It forms the foundation of differentiability and the calculation of derivatives.
### Understanding LimitsA limit describes the behavior of a function; it doesn't necessarily tell you the value of the function at the point. The formal notation used is:
### ExampleConsider the function \( f(x) = \frac{x^2 - 1}{x - 1} \).
Limits are essential for understanding continuous changes and are instrumental in defining concepts such as derivatives and integrals.
### Understanding LimitsA limit describes the behavior of a function; it doesn't necessarily tell you the value of the function at the point. The formal notation used is:
- \( \lim_{x \to c} f(x) = L \)
### ExampleConsider the function \( f(x) = \frac{x^2 - 1}{x - 1} \).
- Substituting \( x = 1 \) gives a division by zero, undefined directly.
- However, by factoring, \( f(x) = x + 1 \) for all \( x eq 1 \).
- The limit as \( x \to 1 \) is \( \lim_{x \to 1} f(x) = 2 \).
Limits are essential for understanding continuous changes and are instrumental in defining concepts such as derivatives and integrals.
Function Differentiability
Function differentiability is an extension of understanding limits and derivatives. A function is said to be differentiable at a point if its derivative exists at that point.
### Requirements for DifferentiabilityDifferentiability implies that:
### ExampleTake the simple function \( f(x) = x^3 \):
Understanding differentiability is crucial as it allows for smooth approximations of functions, which are vital in real-world applications like machine learning, physics simulations, and more.
### Requirements for DifferentiabilityDifferentiability implies that:
- The function is continuous at that point.
- The limit of its derivative, derived from the difference quotient, exists at that point.
### ExampleTake the simple function \( f(x) = x^3 \):
- It's differentiable at all points because \( f'(x) = 3x^2 \) is defined everywhere.
Understanding differentiability is crucial as it allows for smooth approximations of functions, which are vital in real-world applications like machine learning, physics simulations, and more.