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Chapter 2: Problem 30

The given limit is a derivative, but of what function and at what point? \(\lim _{x \rightarrow 3} \frac{x^{3}+x-30}{x-3}\)

Short Answer

Expert verified
The function is \( f(x) = x^3 + x \) at point \( x = 3 \).

Step by step solution

01

Recognize the Limit as a Derivative

The given limit expression \( \lim_{x \rightarrow 3} \frac{x^3 + x - 30}{x-3} \) is in the form of a derivative based on the definition of a derivative at a point: \( \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a} \). In this case, the form suggests finding the derivative of a function \( f(x) \) at the point \( x = 3 \).
02

Identify the Function and Point

Compare the given expression to the standard derivative form. Here, \( a = 3 \), and it seems \( f(x) \) should be \( x^3 + x \) since it would result in the difference quotient collapsing to the given limit form \( \frac{x^3 + x - 30}{x-3} \). We suspect \( f(x) = x^3 + x \) and \( f(3) = 30 \), based on the expression \( x^3 + x - 30 \).
03

Differentiate the Function

Compute the derivative of the function \( f(x) = x^3 + x \). The derivative \( f'(x) \) is found by applying the power rule: \[ f'(x) = \frac{d}{dx}(x^3 + x) = 3x^2 + 1 \].
04

Evaluate the Derivative at the Point

Using the derivative \( f'(x) = 3x^2 + 1 \), evaluate it at \( x = 3 \): \[ f'(3) = 3(3)^2 + 1 = 27 + 1 = 28 \]. Therefore, the function is \( f(x) = x^3 + x \) and the derivative at \( x = 3 \) is 28.

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Key Concepts

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

Limit
The concept of a limit is fundamental in calculus. It helps us understand how a function behaves as it approaches a certain point. When we write \( \lim_{x \rightarrow a} f(x) \), we're describing what happens to the value of \( f(x) \) as \( x \) gets infinitely close to \( a \).
In the context of our exercise, the expression \( \lim_{x \rightarrow 3} \frac{x^{3}+x-30}{x-3} \) is particularly important. It suggests that as \( x \) approaches 3, the value of the expression approaches a certain number. This limit is not just an ordinary limit but also represents the derivative of a certain function at a point, a deeper concept we will explore next.
Derivative
A derivative provides a measure of how a function changes as its input changes. It essentially describes the "slope" of the function at any given point. Mathematically, the derivative of a function \( f(x) \) at a point \( a \) is the limit:
  • \( \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a} \).
In our exercise, we've identified the expression \( \lim_{x \rightarrow 3} \frac{x^{3}+x-30}{x-3} \) as representing the derivative of the function \( f(x) \) at \( x = 3 \). This concept unlocks our ability to determine how the function behaves precisely at that point, revealing insights into the rate at which the function's values are changing.
Power Rule
The power rule is a basic rule of differentiation used to easily find the derivative of polynomial functions. The rule states that for any term \( x^n \), the derivative is \( nx^{n-1} \).
Applying the power rule to our function \( f(x) = x^3 + x \), we find its derivative \( f'(x) \) to be:
  • \( \frac{d}{dx}(x^3) = 3x^2 \)
  • \( \frac{d}{dx}(x) = 1 \)
Thus, the derivative \( f'(x) = 3x^2 + 1 \). This use of the power rule demonstrates how straightforward it can be to find derivatives of polynomial expressions, highlighting its importance as a fundamental tool in calculus.
Function Evaluation
To fully solve the problem, we must evaluate the derivative at a specific point, which illustrates how the derivative behaves at that point.
In our exercise, after differentiating \( f(x) = x^3 + x \) using the power rule, we evaluate its derivative at \( x = 3 \). The result is:
  • \( f'(3) = 3(3)^2 + 1 \)
  • = 27 + 1 \li>= 28
This tells us that at \( x = 3 \), the slope of the tangent to the curve \( y = x^3 + x \) is 28. Function evaluation at specific points is crucial because it gives precise information about the function's behavior at those points.

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Most popular questions from this chapter

Use the definition of the derivative to show that \(D_{x}(\sin 5 x)=5 \cos 5 x\)

Use a CAS to do problems. Draw the graphs of \(f(x)=\cos x-\sin (x / 2)\) and its derivative \(f^{\prime}(x)\) on the interval [0,9] using the same axes. (a) Where on this interval is \(f^{\prime}(x)>0 ?\) (b) Where on this interval is \(f(x)\) increasing as \(x\) increases? (c) Make a conjecture. Experiment with other intervals and other functions to support this conjecture.

Find \(D_{x} y .\) \(y=x^{2} \cos x\)

Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). \(F(x)=\frac{6}{x^{2}+1}\)

Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). \(f(x)=x^{3}+2 x^{2}+1\)
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