/*! 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 51 Find $$ \frac{f(x+h)-f(x)}{h... [FREE SOLUTION] | 91影视

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Chapter 0: Problem 51

Find $$ \frac{f(x+h)-f(x)}{h} \text { and } \frac{f(w)-f(x)}{w-x} $$ Simplify as much as possible. $$ f(x)=3 x^{2}-5 $$

Short Answer

Expert verified
For \(\frac{f(x+h)-f(x)}{h}\), the answer is \(6x + 3h\). For \(\frac{f(w)-f(x)}{w-x}\), the answer is \(3(w + x)\).

Step by step solution

01

Identify the Function

We are given \[ f(x) = 3x^2 - 5 \]This is a quadratic function, which will be used for both differences in the expressions provided.
02

Substitute into First Difference Quotient

First, compute the expression \[ \frac{f(x+h)-f(x)}{h} \]Substitute into the formula: \[ f(x+h) = 3(x+h)^2 - 5 \]Then \[ f(x) = 3x^2 - 5 \]
03

Expand and Simplify the First Difference

Expand\[ f(x+h) = 3(x^2 + 2xh + h^2) - 5 = 3x^2 + 6xh + 3h^2 - 5 \]So \[ \frac{f(x+h) - f(x)}{h} = \frac{(3x^2 + 6xh + 3h^2 - 5) - (3x^2 - 5)}{h} \]Cancel out like terms: \[ = \frac{6xh + 3h^2}{h} \]Which simplifies to: \[ = 6x + 3h \]
04

Substitute into Second Difference Quotient

Now compute the expression \[ \frac{f(w)-f(x)}{w-x} \]Substitute into the formula: \[ f(w) = 3w^2 - 5 \]Then \[ f(x) = 3x^2 - 5 \]
05

Simplify the Second Difference Quotient

So \[ \frac{f(w) - f(x)}{w-x} = \frac{(3w^2 - 5) - (3x^2 - 5)}{w-x} \]Cancel out like terms: \[ = \frac{3w^2 - 3x^2}{w-x} \]Factor the numerator: \[ = \frac{3(w^2 - x^2)}{w-x} \]Recognize a difference of squares: \[ = \frac{3(w-x)(w+x)}{w-x} \]Cancel \[ w - x \] to simplify to: \[ = 3(w + x) \]
06

Simplified Formulas

The simplified forms of the given expressions are:\[ \frac{f(x+h)-f(x)}{h} = 6x + 3h \]\[ \frac{f(w)-f(x)}{w-x} = 3(w + x) \]

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Key Concepts

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

Quadratic Function
A quadratic function is one of the most fundamental types of polynomial functions in algebra, represented by the formula \[ f(x) = ax^2 + bx + c \]. In this case, \( f(x) = 3x^2 - 5 \) is our specific quadratic function.
Here, it's clear that:
  • \( a = 3 \)
  • \( b = 0 \)
  • \( c = -5 \)
Quadratic functions have a distinct parabolic shape when graphed, with open ends either upwards or downwards depending on the sign of \( a \). They are called 'quadratic' because the highest degree of \( x \) is 2, which derives from the Latin word "quadratus" meaning square.
The quadratic function is fundamental in a variety of mathematical contexts, including modeling, solving real-world problems, and understanding the concept of rates of change.
Simplification
Simplification in mathematics means breaking down expressions into their most basic form to make them easier to work with. This is exactly what we do when simplifying the difference quotient.
For the expression \( \frac{f(x+h)-f(x)}{h} \), we start by substituting the definition of the function into the formula. After expanding and combining like terms, you end up canceling terms so that:\[ \frac{(3x^2 + 6xh + 3h^2 - 5) - (3x^2 - 5)}{h} \] becomes:
  • \( = \frac{6xh + 3h^2}{h} \)
  • Which simplifies further to \( = 6x + 3h \)
Similarly, for \( \frac{f(w) - f(x)}{w-x} \), simplification involves recognizing patterns and factoring:\[ \frac{3(w-x)(w+x)}{w-x} \] and then canceling the \( w-x \) terms, giving us the final simplified expression:
  • \( 3(w + x) \)
By practicing these steps, students learn to approach more complex algebraic expressions with confidence.
Difference of Squares
The difference of squares is a specific algebraic form and an important technique in simplifying expressions like the difference quotient. The formula for a difference of squares is:\[ a^2 - b^2 = (a - b)(a + b) \]This formula plays a crucial role in simplifying expressions by transforming them into products of sums and differences.
In the second difference quotient \( \frac{f(w) - f(x)}{w-x} \), we encounter an expression of the form \( w^2 - x^2 \), which is a classic difference of squares.
By applying the formula, it transforms to:\[ 3(w^2 - x^2) = 3(w - x)(w + x) \]The ability to recognize and apply the difference of squares in this context allows students to reduce complex algebraic expressions to simpler forms. This skill not only aids in solving exercises quickly but also enhances analytical skills within mathematics.

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Most popular questions from this chapter

Equations of the form $$ x=A_{1} \sin \omega t+A_{2} \cos \omega t $$ arise in the study of vibrations and other periodic motion. Express the equation $$ x=\sqrt{2} \sin 2 \pi+\sqrt{6} \cos 2 \pi t $$ in the form \(x=A \sin (\omega t+\theta),\) and use a graphing utility to confirm that both equations have the same graph.

An airplane is flying at a constant height of \(3000 \mathrm{ft}\) above water at a speed of \(400 \mathrm{ft} / \mathrm{s}\). The pilot is to release a survival package so that it lands in the water at a sighted point \(P .\) If air resistance is neglected, then the package will follow a parabolic trajectory whose equation relative to the coordinate system in the accompanying figure is $$ y=3000-\frac{g}{2 v^{2}} x^{2} $$ where \(g\) is the acceleration due to gravity and \(v\) is the speed of the airplane. Using \(g=32 \mathrm{ft} / \mathrm{s}^{2},\) find the "line of sight" angle \(\theta,\) to the nearest degree, that will result in the package hitting the target point.

Find a formula for \(f^{-1}(x)\) $$ f(x)=3 / x^{2}, \quad x<0 $$

The number of hours of daylight on a given day at a given point on the Earth's surface depends on the latitude \(\lambda\) of the point, the angle \(\gamma\) through which the Earth has moved in its orbital plane during the time period from the vernal equinox (March \(21),\) and the angle of inclination \(\phi\) of the Earth's axis of rotation measured from ecliptic north \(\left(\phi \approx 23.45^{\circ}\right) .\) The number of hours of daylight \(h\) can be approximated by the formula $$ h=\left\\{\begin{array}{ll}{24,} & {D \geq 1} \\ {12+\frac{2}{15} \sin ^{-1} D,} & {|D|<1} \\ {0,} & {D \leq-1}\end{array}\right. $$ $$ \begin{array}{l}{\text { where }} \\ {\qquad D=\frac{\sin \phi \sin \gamma \tan \lambda}{\sqrt{1-\sin ^{2} \phi \sin ^{2} \gamma}}}\end{array} $$ and \(\sin ^{-1} D\) is in degree measure. Given that Fairbanks, Alaska, is located at a latitude of \(\lambda=65^{\circ} \mathrm{N}\) and also that \(\gamma=90^{\circ}\) on June 20 and \(\gamma=270^{\circ}\) on December \(20,\) approximate (a) the maximum number of daylight hours at Fairbanks to one decimal place (b) the minimum number of daylight hours at Fairbanks to one decimal place.

A soccer player kicks a ball with an initial speed of 14 \(\mathrm{m} / \mathrm{s}\) at an angle \(\theta\) with the horizontal (see the accompanying figure). The ball lands \(18 \mathrm{m}\) down the field. If air resistance is neglected, then the ball will have a parabolic trajectory and the horizontal range \(R\) will be given by $$ R=\frac{v^{2}}{g} \sin 2 \theta $$ where \(v\) is the initial speed of the ball and \(g\) is the acceleration due to gravity. Using \(g=9.8 \mathrm{m} / \mathrm{s}^{2},\) approximate two values of \(\theta,\) to the nearest degree, at which the ball could have been kicked. Which angle results in the shorter time of flight? Why?
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