/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Simplify the quantities using \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Simplify the quantities using \(m(z)=z^{2}\). $$m(z+h)-m(z)$$

Short Answer

Expert verified
The simplified form is \(2zh + h^2\).

Step by step solution

01

Understand the Function

The function given is \(m(z) = z^2\). This means that it takes any input \(z\), and outputs \(z\) squared. This function will be used to transform both \(z+h\) and \(z\) separately.
02

Apply the Function to \(z+h\)

We need to find the function's output when the input is \(z+h\). Using our function \(m(z) = z^2\), we calculate \(m(z+h) = (z+h)^2\).
03

Expand \((z+h)^2\)

Apply the formula for the square of a binomial: \((z+h)^2 = z^2 + 2zh + h^2\).
04

Apply the Function to \(z\)

Now, compute \(m(z) = z^2\). This is simply \(z^2\) since it's directly from the function definition \(m(z) = z^2\).
05

Subtract \(m(z)\) from \(m(z+h)\)

Substitute the expressions obtained: \(m(z+h) - m(z) = (z^2 + 2zh + h^2) - z^2\).
06

Simplify the Expression

Combine like terms to simplify: \(z^2 + 2zh + h^2 - z^2 = 2zh + h^2\). The \(z^2\) terms cancel out.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Function Simplification
Function simplification is a key process in calculus that involves reducing complex expressions into a simpler, more manageable form. The goal is to make calculations easier and to identify patterns or behaviors of a function. In this exercise, we worked with the function \( m(z) = z^2 \). By applying this function to different expressions like \( z+h \), we expand and then simplify the resulting expressions.

To simplify \( m(z+h) - m(z) \), we did the following:
  • Calculated \( m(z+h) = (z+h)^2 \) and then expanded it using the distributive property.
  • Found \( m(z) \) which is simply \( z^2 \).
  • Subtracted the two expressions: \( (z^2 + 2zh + h^2) - z^2 \).
  • Cancelled out the \( z^2 \) terms to get \( 2zh + h^2 \).
The process shows how simplifying functions involves expanding, subtracting, and canceling terms to arrive at a more concise expression. This makes it easier to analyze and work with further in calculus problems.
Binomial Expansion
Binomial expansion is a technique used to expand expressions that are raised to a power, typically seen in the form \((a + b)^n\). In our exercise, we encountered this when transforming \((z+h)^2\).The general formula for expanding a binomial is
  • \((a + b)^2 = a^2 + 2ab + b^2\)
In the specific example of \((z+h)^2\), substituting \(a=z\) and \(b=h\) into the formula gives us:
  • \((z+h)^2 = z^2 + 2zh + h^2\)
This expanded form allows us to see each term clearly, making it easier to perform operations like subtraction. By applying binomial expansion, we transformed the expression into a form that was easily simplified by combining and canceling like terms. Binomial expansion is a fundamental tool in calculus that frequently appears in polynomials and equations involving squares or higher powers.
Difference Quotient
The difference quotient is a crucial concept in calculus, often used to find the derivative of a function. It represents the average rate of change of the function over a small interval and is defined as:
  • \( \frac{f(x+h) - f(x)}{h} \)
In the given problem, while we didn’t compute the full difference quotient, the expression \( m(z+h) - m(z) \) is an integral part of it. This part \( 2zh + h^2 \) would be what you'd prepare before dividing by \( h \) to find the derivative of the function \( m(z) = z^2 \).The simplified form \( 2zh + h^2 \) reveals the behavior of the function as \( h \) approaches zero, and this is how the derivative is fundamentally formed. When \( h \) is very small, the \( h^2 \) term becomes negligible, and the derivative trends toward the linear term, helping us understand the slope of the tangent at any point \( z \). Understanding and calculating the difference quotient is vital for grasping the concept of derivatives and differentiation in calculus.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.