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Use the Monotonicity Theorem to find where the given function is increasing and where it is decreasing. $$ h(z)=\frac{z^{4}}{4}-\frac{4 z^{3}}{6} $$

Short Answer

Expert verified
Function is increasing for \( z > 2 \) and decreasing for \( z < 2 \).

Step by step solution

01

Find the derivative of the function

The first step in applying the Monotonicity Theorem is to find the derivative of the function. Given \( h(z) = \frac{z^4}{4} - \frac{4z^3}{6} \), we need to calculate \( h'(z) \): \[ h'(z) = \frac{d}{dz} \left( \frac{z^4}{4} \right) - \frac{d}{dz} \left( \frac{4z^3}{6} \right) \] Using power rules for derivatives:\[ h'(z) = \frac{4z^3}{4} - \frac{12z^2}{6} = z^3 - 2z^2 \].

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Key Concepts

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

Derivative
In calculus, a derivative gives us important information about how a function changes. When finding the derivative of a function, we track how its output changes as the input changes. This is known as the rate of change. For a function like \( h(z) = \frac{z^4}{4} - \frac{4z^3}{6} \), finding the derivative \( h'(z) \) involves using the power rule. The power rule helps calculate derivatives of functions in the form of \( x^n \), resulting in \( nx^{n-1} \).

Let's break down the derivative of \( h(z) \):
  • The derivative of \( \frac{z^4}{4} \) is \( \frac{4z^3}{4} = z^3 \).
  • The derivative of \( -\frac{4z^3}{6} \) is \( -\frac{12z^2}{6} = -2z^2 \).
This gives us the derivative \( h'(z) = z^3 - 2z^2 \).

When evaluating a derivative, you learn about the function's behavior at any point along its domain. These insights are crucial for determining increasing or decreasing trends, which lead us to understand thoroughly how the function behaves.
Increasing and Decreasing Functions
Understanding when a function is increasing or decreasing is key to analyzing its behavior. A function is increasing on intervals where its derivative is positive, and decreasing where its derivative is negative.

The critical points of a function occur where its derivative \( h'(z) \) is zero or undefined. Solving \( h'(z) = z^3 - 2z^2 = 0 \) will pinpoint the function's critical points. We can factor it as \( z^2(z - 2) = 0 \), yielding solutions at \( z = 0 \) and \( z = 2 \).

We further analyze the intervals around these critical points to determine where the function is increasing or decreasing:
  • For \( z < 0 \), test \( h'(z) \) to check if it is positive or negative.
  • For \( 0 < z < 2 \), test again to see function's behavior.
  • For \( z > 2 \), another test to reveal increase or decrease.
After testing, if \( h'(z) > 0 \), the function is increasing; if \( h'(z) < 0 \), it's decreasing. This practical examination lets us fully understand the cycle of changes in the function.
Calculus Problem-Solving
Calculus problem-solving equips us with the tools needed to tackle various mathematical puzzles involving rates of change and accumulations. The Monotonicity Theorem is a powerful tool in this context, allowing us to identify where a function shifts from increasing to decreasing or vice versa.

The process often starts with differential calculus, which utilizes derivatives to analyze function behavior.

Let's recap how to solve problems like these:
  • Take the derivative of the function using appropriate rules (like the power rule).
  • Solve for critical points by setting the derivative equal to zero.
  • Determine intervals of increase or decrease by evaluating the sign of the derivative around the critical points.
These steps form the basis for effectively applying calculus to real-world problems, helping you understand trends and behaviors in complex systems. Mastering these techniques can provide clarity and insight into various functional relationships, whether in academia or practical applications.

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