/*! 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 66 $$ \begin{array}{l} \text { ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

$$ \begin{array}{l} \text { For each function, find and simplify }\\\ \frac{f(x+h)-f(x)}{h} . \quad(\text { Assume } h \neq 0 .) \end{array} $$ $$ f(x)=3 x^{2}-5 x+2 $$

Short Answer

Expert verified
The simplified expression is \( 6x + 3h - 5 \).

Step by step solution

01

Understand the Function

We are given a quadratic function \( f(x) = 3x^2 - 5x + 2 \). We need to find the expression \( \frac{f(x+h)-f(x)}{h} \). This expression is crucial for understanding the rate of change of the function and is related to the derivative.
02

Substitute \( x+h \) into the Function

Substitute \( x+h \) into the function \( f(x) \):\[ f(x+h) = 3(x+h)^2 - 5(x+h) + 2 \]Simplify this expression by expanding the binomials.
03

Expand and Simplify \( f(x+h) \)

First expand \((x+h)^2\) to get \(x^2 + 2xh + h^2\). Then substitute:\[ f(x+h) = 3(x^2 + 2xh + h^2) - 5(x+h) + 2 \]Simplify further:\[ = 3x^2 + 6xh + 3h^2 - 5x - 5h + 2 \].
04

Find \( f(x+h) - f(x) \)

Calculate \( f(x+h) - f(x) \) by substituting in the expressions for \( f(x+h) \) and \( f(x) \):\[ f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 5x - 5h + 2) - (3x^2 - 5x + 2) \]Simplify:\[ = 6xh + 3h^2 - 5h \].
05

Divide by \( h \) and Simplify

Now divide the expression from Step 4 by \( h \):\[ \frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 - 5h}{h} \]Factor \( h \) out from the numerator:\[ = \frac{h(6x + 3h - 5)}{h} \]Cancel out \( h \) to simplify:\[ = 6x + 3h - 5 \].
06

Conclusion

The expression \( \frac{f(x+h)-f(x)}{h} \) simplifies to \( 6x + 3h - 5 \). This shows the average rate of change of the function around \( x \) as \( h \) approaches 0.

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.

Quadratic Function
A quadratic function is a polynomial function of degree two, generally expressed in the form \( f(x) = ax^2 + bx + c \). It is characterized by the presence of the term \( x^2 \), which imparts a parabolic shape to the graph of the function. In our specific problem, the quadratic function given is \( f(x) = 3x^2 - 5x + 2 \).Key features of quadratic functions include:
  • The vertex, which is the highest or lowest point of the parabola depending on the sign of \( a \).
  • The axis of symmetry, a vertical line that passes through the vertex and divides the parabola into two mirror images.
  • The direction of the parabola, which opens upwards if \( a > 0 \) and downwards if \( a < 0 \).
Quadratic functions appear frequently in various fields like physics, engineering, and economics because they model relationships where variables change under constraints, describing paths of projectiles, profit optimization, and more.
Rate of Change
The rate of change is a fundamental concept that describes how a function's output changes with respect to changes in the input. For a function \( f(x) \), the rate of change over an interval is determined by the difference quotient, which is expressed as \( \frac{f(x+h) - f(x)}{h} \).In the context of the given quadratic function, the simplified difference quotient \( 6x + 3h - 5 \) is used. This expression tells us how the output of the quadratic changes as \( h \) changes, providing an average rate of change over a small interval around \( x \).Understanding the rate of change:
  • Helps us anticipate the behavior of the function over an interval.
  • Indicates how fast or slow the function's value is increasing or decreasing.
The rate of change becomes especially valuable when examining trends and patterns or making predictions based on the function's graph.
Derivative
The derivative of a function is a central concept in calculus that represents the instantaneous rate of change of the function with respect to one of its variables. When \( h \) approaches zero, the difference quotient \( \frac{f(x+h) - f(x)}{h} \) transforms into the derivative.In relation to our quadratic function, as \( h \to 0 \), the expression \( 6x + 3h - 5 \) simplifies to \( 6x - 5 \). This is the derivative of the quadratic function \( f(x) = 3x^2 - 5x + 2 \).Key insights from derivatives:
  • They provide critical information about the function's increasing or decreasing behavior at a specific point.
  • Derivatives allow us to find the slope of the tangent to the function's graph at any point \( x \).
  • By identifying where the derivative equals zero, we can locate the function's critical points, which help determine maxima, minima, or inflection points.
Mastering the derivative concept is essential for tackling more complex problems in calculus and making informed decisions in applied mathematics settings.

One App. One Place for Learning.

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

Get started for free

Most popular questions from this chapter

$$ \begin{array}{l} \text { For each function, find and simplify }\\\ \frac{f(x+h)-f(x)}{h} . \quad(\text { Assume } h \neq 0 .) \end{array} $$ $$ \begin{array}{l} f(x)=x^{4} \\ \text { [Hint: Use }(x+h)^{4}=x^{4}+4 x^{3} h+6 x^{2} h^{2}+ \\ \left.4 x h^{3}+h^{4} .\right] \end{array} $$

BUSINESS: Straight-Line Depreciation Straight-line depreciation is a method for estimating the value of an asset (such as a piece of machinery) as it loses value ( \({ }^{\prime \prime}\) depreciates" \()\) through use. Given the original price of an asset, its useful lifetime, and its scrap value (its value at the end of its useful lifetime), the value of the asset after \(t\) years is given by the formula: $$ \begin{aligned} \text { Value }=(\text { Price })-&\left(\frac{(\text { Price })-(\text { Scrap value })}{(\text { Useful lifetime })}\right) \cdot t \\ & \text { for } 0 \leq t \leq(\text { Useful lifetime }) \end{aligned} $$ a. A newspaper buys a printing press for $$\$ 800,000$$ and estimates its useful life to be 20 years, after which its scrap value will be $$\$ 60,000$$. Use the formula above Exercise 63 to find a formula for the value \(V\) of the press after \(t\) years, for \(0 \leq t \leq 20\) b. Use your formula to find the value of the press after 10 years. c. Graph the function found in part (a) on a graphing calculator on the window \([0,20]\) by \([0,800,000] .\) [Hint: Use \(x\) instead of \(t\).]

$$ \text { How do the graphs of } f(x) \text { and } f(x)+10 \text { differ? } $$

GENERAL: Stopping Distance A car traveling at speed \(v\) miles per hour on a dry road should be able to come to a full stop in a distance of $$ D(v)=0.055 v^{2}+1.1 v \text { feet } $$ Find the stopping distance required for a car traveling at: \(60 \mathrm{mph}\).

Smoking and Income Based on a recent study, the probability that someone is a smoker decreases with the person's income. If someone's family income is \(x\) thousand dollars, then the probability (expressed as a percentage) that the person smokes is approximately \(y=-0.31 x+40\) (for \(10 \leq x \leq 100)\) a. Graph this line on the window \([0,100]\) by \([0,50]\). b. What is the probability that a person with a family income of $$\$ 40,000$$ is a smoker? [Hint: Since \(x\) is in thousands of dollars, what \(x\) -value corresponds to $$\$ 40,000 ?]$$ c. What is the probability that a person with a family income of $$\$ 70,000$$ is a smoker? Round your answers to the nearest percent.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.