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Chapter 2: Problem 36

\(31-36\) Each limit represents the derivative of some function \(f\) at some number a. State such an \(f\) and a in each case. $$\lim _{t \rightarrow 1} \frac{t^{4}+t-2}{t-1}$$

Short Answer

Expert verified
The function is \( f(t) = t^4 + t - 2 \) and \( a = 1 \).

Step by step solution

01

Identify the Structure of the Limit

The given limit \( \lim_{t \rightarrow 1} \frac{t^{4} + t - 2}{t - 1} \) resembles the definition of the derivative, which is \( \lim_{t \rightarrow a} \frac{f(t) - f(a)}{t-a} \). To find \( f(t) \) and \( a \), note that \( t-1 \) in the denominator suggests \( a = 1 \). The numerator \( t^{4}+t-2 \) can be viewed as \( f(t) - f(1) \).
02

Determine the Function \( f(t) \)

Start by setting \( f(1) \) to be a term that balances the expression obtained when substituting \( t = 1 \) into \( f(t) \). Substitute \( t = 1 \) into \( t^4 + t - 2 \) to get \( 1^4 + 1 - 2 = 0 \). Therefore, \( f(1) = 0 \). Hence, \( f(t) = t^4 + t - 2 \).
03

Verify \( f(t) \) and \( a \)

Since we have defined \( f(t) = t^4 + t - 2 \), check that \( f(1) = 0 \) satisfies the condition. Substitute \( t = 1 \) into \( f(t) \) to get \( 1^4 + 1 - 2 = 0 \). The structure of the limit then confirms that it is indeed in the form for \( \frac{f(t) - f(1)}{t-1} \).
04

Conclusion

The function \( f(t) = t^4 + t - 2 \) and the number \( a = 1 \) satisfy the given limit as the derivative \( \lim_{t \rightarrow 1} \frac{f(t) - f(1)}{t-1} \).

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Key Concepts

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

Limits
A limit is the value that a function \(f(t)\) approaches as the variable \(t\) gets closer to a particular point. In calculus, limits are used to define derivatives, which measure how a function changes.

In the exercise above, the limit \(\lim_{t \rightarrow 1} \frac{t^{4}+t-2}{t-1}\) indicates the behavior of the function as \(t\) approaches 1.
  • The denominator \(t - 1\) vanishing at \(t = 1\) suggests a special behavior or point of interest at 1.
  • The presence of \(t - 1\) in the denominator hints at the derivative of a function at \(a = 1\).
Exploring limits is a key step in understanding the derivative and analyzing how functions work.
Numerical Differentiation
Numerical differentiation is the process of estimating the derivative of a function using limits. This is particularly useful when dealing with functions that are not simple polynomials.

Consider our function \(f(t) = t^4 + t - 2\). The derivative reflects the rate of change of this function. Understanding the change of \(f\) around a specific point \(a\) (like \(a = 1\)) allows us to predict how small alterations in \(t\) affect \(f(t)\).
  • By using the limit structure \(\lim_{t \rightarrow 1} \frac{f(t) - f(1)}{t-1}\), we can numerically derive the function's behavior at \(t = 1\).
  • This demonstrates the fundamental definition of the derivative, offering insights into the function's behavior near any chosen point.
Through numerical differentiation, we gain valuable insights into contexts where symbolic differentiation could be cumbersome.
Function Behavior
Analyzing the behavior of a function involves understanding its changes and tendencies over its domain. This is often achieved through examining derivatives at various points.

For the given function \(f(t) = t^4 + t - 2\), we are interested in what happens when \(t\) nears 1.
  • First, by determining \(f(t)\)'s derivative at \(t = 1\), we gauge how rapidly the function value changes.
  • Second, understanding the derivative's sign indicates the function is increasing or decreasing; a positive derivative suggests increasing behavior, while a negative derivative suggests decreasing.
Deeply understanding function behavior unlocks further comprehension of a function's full narrative.
Calculus Problem Solving
Calculus problem solving hinges on transforming real-world problems into mathematical equations that can be analyzed and solved. Derivatives play a central role in this process.

For solving the problem \(\lim _{t \rightarrow 1} \frac{t^{4}+t-2}{t-1}\), we must first recognize it as a derivative.
  • Identifying and defining \(f(t)\) and \(a = 1\) helped structure and simplify the problem.
  • Knowing that the limit's format aligns with the derivative provides a clear path to the solution.
  • From there, use algebraic techniques to verify \(f(t) = t^4 + t - 2\) satisfies \(f(1) = 0\) which confirms the problem's setup.
By mastering these steps, students can tackle complex calculus problems methodically and accurately.

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Most popular questions from this chapter

Let \(P(t)\) be the percentage of Americans under the age of 18 at time \(t .\) The table gives values of this function in census years from 1950 to \(2000.\) $$\begin{array}{|c|c|c|c|}\hline t & {P(t)} & {t} & {P(t)} \\ \hline 1950 & {31.1} & {1980} & {28.0} \\ {1960} & {35.7} & {1990} & {25.7} \\ {1970} & {34.0} & {2000} & {25.7} \\ \hline\end{array}$$ (a) What is the meaning of \(\mathrm{P}^{\prime}(\mathrm{t}) ?\) What are its units? (b) Construct a table of estimated values for \(\mathrm{P}^{\prime}(\mathrm{t}) .\) (c) Graph \(\mathrm{P}\) and \(\mathrm{P}^{\prime}\) (d) How would it be possible to get more accurate values for \(\mathrm{P}^{\prime}(\mathrm{t}) ?\)

If $$\lim _{x \rightarrow 1} \frac{f(x)-8}{x-1}=10,$$ find $$\lim _{x \rightarrow 1} f(x).$$

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. \(g(\mathrm{t})=\frac{1}{\sqrt{\mathrm{t}}}\)

$$ \lim _{x \rightarrow-\infty} \frac{1}{x}=0 $$

A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after t minutes. When the data in the table are graphed, the slope of the tangent line represents the heart rate in beats per minute. $$\begin{array}{|c|c|c|c|c|c|}\hline t(\min ) & {36} & {38} & {40} & {42} & {44} \\ \hline \text { Heartbeats } & {2530} & {2661} & {2806} & {2948} & {3080} \\ \hline\end{array}$$ The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the patient's heart rate after 42 minutes using the secant line between the points with the given values of \(t\) . $$\begin{array}{ll}{\text { (a) } t=36} & {\text { and } t=42 \quad \text { (b) } t=38 \text { and } t=42} \\ {\text { (c) } t=40 \text { and } t=42} & {\text { (d) } t=42 \text { and } t=44}\end{array}$$ What are your conclusions?
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