Chapter 1: Problem 10
For \(F(t)=4 t^{3}\), find and simplify \([F(a+h)-F(a)] / h\).
Short Answer
Expert verified
\([F(a+h)-F(a)] / h\) simplifies to \(12a^2 + 12ah + 4h^2\).
Step by step solution
01
Write the Expression for F(t)
We are given the function \( F(t) = 4t^3 \). We need to find \( [F(a+h) - F(a)] / h \).
02
Find F(a+h)
Substitute \( a+h \) into \( t \) in the function: \[F(a+h) = 4(a+h)^3\]
03
Expand (a+h)^3
Use the binomial theorem to expand \((a+h)^3\):\[(a+h)^3 = a^3 + 3a^2h + 3ah^2 + h^3\]
04
Calculate F(a+h)
Substitute the expansion back into our expression for \(F(a+h)\): \[F(a+h) = 4(a^3 + 3a^2h + 3ah^2 + h^3) = 4a^3 + 12a^2h + 12ah^2 + 4h^3\]
05
Calculate F(a)
Substitute \(a\) in the function:\[F(a) = 4a^3\]
06
Substitute into the Difference Quotient
Substitute \(F(a+h)\) and \(F(a)\) into the difference quotient:\[[F(a+h) - F(a)] / h = \frac{(4a^3 + 12a^2h + 12ah^2 + 4h^3) - 4a^3}{h}\]
07
Simplify the Numerator
Cancel \(4a^3\) terms and simplify:\[12a^2h + 12ah^2 + 4h^3\]
08
Factor Out h from the Numerator
Factor out \(h\) from the numerator:\[h(12a^2 + 12ah + 4h^2)\]
09
Cancel h in the Expression
Cancel the \(h\) in the numerator and denominator:\[12a^2 + 12ah + 4h^2\]
10
Write the Simplified Expression
Thus, the simplified expression for \([F(a+h)-F(a)] / h\) is:\[12a^2 + 12ah + 4h^2\]
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.
Difference Quotient
The difference quotient is a fundamental tool in calculus used to find the derivative of a function at a particular point. It essentially measures the average rate of change of the function over a small interval. We express it mathematically as \( \frac{f(x+h) - f(x)}{h} \). This formula evaluates how the output of a function changes as the input is increased by a small amount \(h\). It serves as the precursor to the derivative, which is the rate of change at a single instant.
In the exercise, the function \(F(t) = 4t^3\) requires us to find the simplified form of \([F(a+h) - F(a)] / h\). By substituting and simplifying, one determines how the function \(F(t)\) behaves as its variable \(t\) slightly increases from \(a\) to \(a+h\).
In the exercise, the function \(F(t) = 4t^3\) requires us to find the simplified form of \([F(a+h) - F(a)] / h\). By substituting and simplifying, one determines how the function \(F(t)\) behaves as its variable \(t\) slightly increases from \(a\) to \(a+h\).
- The numerator \([F(a+h) - F(a)]\) represents the change in the function's value between \(a\) and \(a+h\).
- Dividing by \(h\) gives us the average rate of change over that interval.
- As \(h\) approaches zero, the difference quotient gives us the derivative of the function.
Binomial Theorem
The binomial theorem provides a method to expand expressions that are raised to a power, especially of the form \((a + b)^n\). An expansion using the binomial theorem is written as:\[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]This formula allows us to expand polynomial expressions easily, which is crucial in calculus and algebra.
In our problem, the binomial theorem is applied to expand \((a+h)^3\). It shows how we can break down complex expressions into simpler terms:
In our problem, the binomial theorem is applied to expand \((a+h)^3\). It shows how we can break down complex expressions into simpler terms:
- \(a^3\) - The term when only \(a\) appears from the binomial term.
- \(3a^2h\) - The term accounting the combination of \(a\) appearing twice and \(h\) appearing once.
- \(3ah^2\) - Illustrates \(a\) once and \(h\) twice.
- \(h^3\) - Represents \(h\) raised to the third power exclusively.
Polynomial Functions
Polynomial functions are expressions involving variables raised to whole number powers. They can be simple, like a quadratic \(ax^2 + bx + c\), or more complex like the cubic function found in our exercise, \(4t^3\). They are characterized by their smooth and continuous curves.
Polynomials are highly useful in differential calculus because their derivatives are straightforward to calculate. In fact, the power rule - which simplifies the differentiation of polynomials - states that the derivative of \(x^n\) is \(nx^{n-1}\).
In the problem's context, the polynomial \(4t^3\) involves:
Polynomials are highly useful in differential calculus because their derivatives are straightforward to calculate. In fact, the power rule - which simplifies the differentiation of polynomials - states that the derivative of \(x^n\) is \(nx^{n-1}\).
In the problem's context, the polynomial \(4t^3\) involves:
- The highest power is 3, indicating a cubic polynomial.
- As per binomial expansion and simplification, the function transforms and simplifies in its difference quotient form.
- Polynomials remain a cornerstone in calculus due to their predictable behavior and wide application.