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Chapter 4: Problem 28

\(23-36=\) Find the critical numbers of the function. $$g(t)=|3 t-4|$$

Short Answer

Expert verified
The critical number is \(t = \frac{4}{3}\).

Step by step solution

01

Understand the Concept

A critical number of a function is a value of the variable in the domain of the function where the derivative is zero or undefined. For the function \(g(t) = |3t - 4|\), we need to first find the derivative and determine where it is zero or undefined.
02

Express Absolute Value as a Piecewise Function

Rewrite the absolute value function \(g(t) = |3t - 4|\) as a piecewise function:- If \(3t - 4 \geq 0\), then \(g(t) = 3t - 4\).- If \(3t - 4 < 0\), then \(g(t) = -(3t - 4) = -3t + 4\).
03

Determine Intervals

Find the point where \(3t - 4 = 0\). Solve this equation:\[3t - 4 = 0 \implies t = \frac{4}{3}\].So, the function is piecewise and changes at \(t = \frac{4}{3}\).
04

Find Derivatives on Each Interval

Calculate the derivative for each piece of the function:- On the interval \(t < \frac{4}{3}\), where \(g(t) = -3t + 4\), the derivative \(g'(t) = -3\).- On the interval \(t > \frac{4}{3}\), where \(g(t) = 3t - 4\), the derivative \(g'(t) = 3\).
05

Examine Derivative at Critical Point

Since the derivative changes from \(-3\) to \(3\) at \(t = \frac{4}{3}\), this means that the derivative is undefined at \(t = \frac{4}{3}\). This point is a critical number of the function.
06

Conclusion

The critical number is the point where the derivative changes from one value to another, which happens at \(t = \frac{4}{3}\), because the derivative is not continuous at this point.

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Key Concepts

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

Absolute Value Function
The absolute value function is a vital concept in algebra and calculus. It's defined as the distance of a number from zero on the number line, without considering direction. This means it always outputs a non-negative value.

For instance, the absolute value of both \(3\) and \(-3\) is \(3\). In mathematical terms, the absolute value of a number \(x\) is written as \(|x|\).
  • If \(x \geq 0\), then \(|x| = x\).

  • If \(x < 0\), then \(|x| = -x\).
Understanding the absolute value function is crucial for solving equations or inequalities that involve absolute values. When used within other functions, like in our case with \(g(t) = |3t - 4|\), the function can be approached by expressing it as a piecewise function.
Piecewise Functions
Piecewise functions are mathematical expressions defined by multiple sub-functions, each applying to a certain interval of the main function's domain. These functions break down into simpler pieces, which are defined over specific ranges, or intervals, of the variable.

In the example of \(g(t) = |3t - 4|\), the absolute value expression is rewritten as:
  • \(g(t) = 3t - 4\) for \(3t - 4 \geq 0\)

  • \(g(t) = -3t + 4\) for \(3t - 4 < 0\)
By identifying the point where the pieces change, which occurs when the expression inside the absolute value equals zero, we find \([t = \frac{4}{3}]\). This transition point helps us divide the function into its respective intervals, making it easier to analyze and work with each piece separately.
Derivatives
Derivatives represent one of the fundamental concepts of calculus and measure how a function changes as its input changes. It's akin to calculating the "instantaneous rate of change" or the slope of the tangent line at a particular point on the function.

For the piecewise function derived from \(g(t) = |3t - 4|\), derivatives offer insights into how each segment behaves:
  • Derivative of \(g(t) = -3t + 4\) when \(t < \frac{4}{3}\) is \(g'(t) = -3\).

  • Derivative of \(g(t) = 3t - 4\) when \(t > \frac{4}{3}\) is \(g'(t) = 3\).
The derivative helps uncover critical points, which are where the derivative is zero or undefined. In our piecewise function, the derivative shifts values as the piece changes, highlighting the point of non-differentiability at \(t = \frac{4}{3}\).
Calculus
Calculus is the mathematical study of continuous change and offers tools like derivatives and integrals to work with functions that depict such changes. It plays a critical role in exploring the concepts of motion and growth in natural and physical sciences.

With functions like \(g(t) = |3t - 4|\), calculus allows us to:
  • Understand function behavior through derivatives.
  • Investigate critical points and analyze their significance.
  • Explore how functions transform across intervals using piecewise expressions.
By applying calculus principles, we gain a better understanding of the function's critical numbers and how it behaves differently over distinct intervals. The crucial point in the given function was at \(t = \frac{4}{3}\), revealing where the function's rate of change isn't continuous.

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Most popular questions from this chapter

Suppose \(f\) is differentiable on an interval \(I\) and \(f^{\prime}(x)>0\) for all numbers \(x\) in \(I\) except for a single number \(c .\) Prove that \(f\) is increasing on the entire interval \(I.\)

A stone is dropped from the upper observation deck (the Space Deck) of the CN Tower, 450 \(\mathrm{m}\) above the ground. (a) Find the distance of the stone above ground level at time \(t .\) (b) How long does it take the stone to reach the ground? (c) With what velocity does it strike the ground? (d) If the stone is thrown downward with a speed of \(5 \mathrm{m} / \mathrm{s},\) how long does it take to reach the ground?

A stone was dropped off a cliff and hit the ground with a speed of 120 \(\mathrm{ft} / \mathrm{s} .\) What is the height of the cliff?

\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(x)=3 x^{4}-4 x^{3}-12 x^{2}+1, \quad[-2,3]$$

\(53-56\) (a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places. (b) Use calculus to find the exact maximum and minimum values. $$f(x)=x^{5}-x^{3}+2, \quad-1 \leqslant x \leqslant 1$$
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