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Approximate \(f^{\prime}(1)\) by considering the difference quotient $$ \frac{f(1+h)-f(1)}{h} $$ for values of \(h\) near 0 , and then find the exact value of \(f^{\prime}(1)\) by differentiating. $$ f(x)=x^{3}-3 x+1 $$

Short Answer

Expert verified
The approximate and exact derivative at \( x=1 \) is 0.

Step by step solution

01

Define the function and approximate derivative

We are given the function \( f(x) = x^3 - 3x + 1 \). To approximate \( f'(1) \), use the difference quotient formula: \[ \frac{f(1+h) - f(1)}{h} \] for small values of \( h \).
02

Calculate \( f(1) \) and \( f(1+h) \)

Substitute \( x = 1 \) into the function: \( f(1) = 1^3 - 3(1) + 1 = -1 \). Now, substituting \( x = 1 + h \) into the function gives \( f(1+h) = (1+h)^3 - 3(1+h) + 1 \).
03

Expand \( f(1+h) \)

Expand the expression for \( f(1+h) \):\[(1+h)^3 = 1 + 3h + 3h^2 + h^3\]So, \[ f(1+h) = 1 + 3h + 3h^2 + h^3 - 3 - 3h + 1 = h^3 + 3h^2 + 0h - 1 \].
04

Substitute into the difference quotient

Substitute \( f(1) \) and \( f(1+h) \) into the difference quotient:\[ \frac{(h^3 + 3h^2 - 1) - (-1)}{h} = \frac{h^3 + 3h^2}{h} = h^2 + 3h \].
05

Let \( h \to 0 \) to find the limit

Evaluate the limit as \( h \to 0 \) in \( h^2 + 3h \), which results in \( 0 + 0 = 0 \). Thus, the approximate value of \( f'(1) \) using the difference quotient is 0.
06

Differentiate \( f(x) \) exactly and evaluate \( f'(1) \)

Differentiate \( f(x) = x^3 - 3x + 1 \) using power rule: \( f'(x) = 3x^2 - 3 \). Evaluate this at \( x=1 \):\[ f'(1) = 3(1)^2 - 3 = 3 - 3 = 0 \].
07

Confirm the exact derivative matches the approximation

Both the approximate and exact calculations of \( f'(1) \) resulted in 0, confirming the solution is consistent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference Quotient
The difference quotient is a crucial concept when approximating the derivative of a function. Essentially, it measures the average rate of change of the function over a small interval. For a function \( f(x) \), the difference quotient is expressed as:
  • \( \frac{f(x+h) - f(x)}{h} \)
Here, \( h \) is a small increment in \( x \), and by making \( h \) very small, the difference quotient approaches the derivative of the function at that point.
For the given function \( f(x) = x^3 - 3x + 1 \), we applied this by considering \( f(1+h) - f(1) \). This led to simplifications resulting in \( h^2 + 3h \).
As \( h \) approaches zero, the approximation becomes more accurate, which is an essential aspect of calculus. This approximation helps students get a practical feel of how derivatives function at specific points.
Limit Evaluation
The process of finding the limit of a function as a variable approaches a particular value is called limit evaluation. It is the backbone of finding derivatives in calculus. After simplifying the difference quotient expression to \( h^2 + 3h \), the next step was limit evaluation:
  • We consider the limit as \( h \to 0 \).
For the expression \( h^2 + 3h \), plugging in \( h = 0 \) gives 0. Thus, limit evaluation removes the dependence on \( h \) and ensures what remains is the exact derivative value. This step is necessary because the definition of the derivative is the limit of the difference quotient as \( h \) approaches zero.
Power Rule
The power rule is a quick and efficient method for finding the derivative of functions with terms like \( x^n \). The rule states that:
  • If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).

This was applied to differentiate the given polynomial function \( f(x) = x^3 - 3x + 1 \). By the power rule:
  • The derivative of \( x^3 \) is \( 3x^2 \).
  • For \(-3x\), treating it as \(-3x^1\), the derivative is \(-3 \cdot 1x^{1-1} = -3 \).
The power rule simplifies this process significantly, allowing to quickly find \( f'(x) = 3x^2 - 3 \). Calculating the derivative at \( x = 1 \) confirmed our limit evaluation, showing that both approaches yield \( f'(1) = 0 \). This comprehensive agreement highlights the power rule as a reliable tool for derivative calculations.

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