/*! 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 52 Find $$ \frac{f(x+h)-f(x)}{h... [FREE SOLUTION] | 91Ó°ÊÓ

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Find $$ \frac{f(x+h)-f(x)}{h} \text { and } \frac{f(w)-f(x)}{w-x} $$ Simplify as much as possible. $$ f(x)=x^{2}+6 x $$

Short Answer

Expert verified
\( \frac{f(x+h) - f(x)}{h} = 2x + h + 6 \) and \( \frac{f(w) - f(x)}{w-x} = w+x+6 \)

Step by step solution

01

Define the function increments

We start with the function \( f(x) = x^2 + 6x \). We need to find the increment of this function when we increase \( x \) by \( h \) and by \( w-x \).
02

Compute \( f(x+h) \)

Substitute \( x + h \) into the function:\[f(x+h) = (x+h)^2 + 6(x+h)\]Expanding this, we get:\[f(x+h) = x^2 + 2xh + h^2 + 6x + 6h\]
03

Find \( \frac{f(x+h) - f(x)}{h} \)

We compute the difference \( f(x+h) - f(x) \):\[f(x+h) - f(x) = (x^2 + 2xh + h^2 + 6x + 6h) - (x^2 + 6x)\]This simplifies to:\[2xh + h^2 + 6h\]Divide by \( h \):\[\frac{f(x+h) - f(x)}{h} = 2x + h + 6\]
04

Compute \( f(w) \)

Substitute \( w \) into the function:\[f(w) = w^2 + 6w\]
05

Find \( \frac{f(w) - f(x)}{w-x} \)

We compute the difference \( f(w) - f(x) \):\[f(w) - f(x) = (w^2 + 6w) - (x^2 + 6x)\]This simplifies to:\[w^2 - x^2 + 6w - 6x\]Use the formula \( a^2 - b^2 = (a-b)(a+b) \):\[(w-x)(w+x) + 6(w-x)\]Factor out \( w-x \):\[(w-x)(w+x+6)\]Divide by \( w-x \):\[\frac{f(w) - f(x)}{w-x} = w+x+6\]

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Key Concepts

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

Difference Quotient
The difference quotient is a fundamental concept in calculus, which helps us understand how a function changes as its input changes. It's essentially the slope of the secant line connecting two points on the graph of a function. In simpler terms, it measures how much the function's output changes for a small change in input.
This is calculated as \( \frac{f(x+h) - f(x)}{h} \), where \( h \) is a small increment in the input \( x \). For our specific polynomial function \( f(x) = x^2 + 6x \), this difference quotient translates to the expression \( 2x + h + 6 \) when simplified. Here, we expanded \( f(x+h) \) by substituting \( x+h \) into the function and simplifying the algebraic expression.
  • The term \( 2x \) indicates the change in the first term \( x^2 \), reflecting twice the original input \( x \).
  • The term \( h \) shows the impact of the increment itself.
  • The constant \( 6 \) derives from the linear term \( 6x \), which remains stable except for the addition of \( 6h \).
This illustrates how derivatives are formed through limits of the difference quotient as \( h \) approaches zero.
Polynomial Functions
A polynomial function is an expression constructed from variables (like \( x \)) and coefficients using only the operations of addition, subtraction, multiplication, and non-negative integer exponents. Our exercise uses the polynomial \( f(x) = x^2 + 6x \). This is a simple second-degree polynomial, where \( x^2 \) represents a quadratic term and \( 6x \) is a linear term.
  • The quadratic term \( x^2 \) affects the shape of the graph, making it a parabola opening upwards.
  • The linear term \( 6x \) modifies the slope and the position of the parabola by affecting the graph's steepness and direction.
  • The absence of a constant term (such as \( c \) in a general polynomial \( ax^2 + bx + c \)) focuses the roots of the polynomial at \( x=0 \) since there is no vertical translation.
Understanding how to work with these elements is crucial in solving differential or integral problems involving more complex polynomial expressions.
Limits
Limits are the core concept in calculus that describe the behavior of a function as the input approaches a certain value. They help us formalize the concept of motion and change. For the calculus operation involving the difference quotient, limits help us transition from average rates of change to instantaneous rates of change, which is essentially the derivative.
In our exercise, the limit concept is embodied when we simplify \( \frac{f(x+h) - f(x)}{h} = 2x + h + 6 \). The focus here is on the behavior of this expression as \( h \) approaches zero. As we take the limit \( h \to 0 \), the term \( h \) drops away, leaving us with \( 2x + 6 \), which is the derivative of \( f(x) = x^2 + 6x \).
  • Limits allow us to "zoom in" on the function's behavior at a specific point.
  • This is crucial for understanding the function's instantaneous rate of change (its derivative).
  • Conceptualizing the limit constructively simplifies many seemingly impossible tasks, like calculating motion or growth at a precise moment.
By grasping limits, students can unlock the power of calculus to model and solve real-world problems effectively.

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Most popular questions from this chapter

Let \(f(x)=2 x^{3}+5 x+3 .\) Find \(x\) if \(f^{-1}(x)=1\)

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.

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Prove: A one-to-one function \(f\) cannot have two different inverses.

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