/*! 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 9 If \(f(x)=x^{3}+3 x^{2}-189 x+1,... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 9

If \(f(x)=x^{3}+3 x^{2}-189 x+1,\) find analytically all values of \(x\) for which \(f^{\prime}(x)=0\). (Enter your answer as a comma separated list of numbers, e.g., -1,0,2 ). \(x=\) __________

Short Answer

Expert verified
7, -9

Step by step solution

01

- Find the derivative of the function

To find the critical points where the derivative is zero, first determine the derivative of the function. Given: \(f(x) = x^{3} + 3x^{2} - 189x + 1\)The derivative is found using basic differentiation rules:\[f^{\flat}(x) = \frac{d}{dx}(x^{3} + 3x^{2} - 189x + 1)\]\[f^{\flat}(x) = 3x^{2} + 6x - 189\]
02

- Set the derivative equal to zero

To find the critical points, set the derivative equal to zero:\[3x^{2} + 6x - 189 = 0\]
03

- Simplify the equation

Divide the entire equation by 3 to simplify:\[x^{2} + 2x - 63 = 0\]
04

- Solve the quadratic equation

Solve the quadratic equation \(x^{2} + 2x - 63 = 0\) using the quadratic formula \(x =\frac{-b \backsime; \backslash \backslash sqrt{b^2-4ac}}{2a}\). Here, \(a = 1\), \(b = 2\), and \(c = -63\).\[x =\frac{-2 \backsime; \backslash \backslash sqrt{(2)^2-4(1)(-63)}}{2(1)}\]\[x = \frac{-2 \backsime; \backslash \backslash sqrt{4 + 252}}{2}\]\[x =\frac{-2 \backsime; \backslash \backslash sqrt{256}}{2}\]\[x =\frac{-2 \backsime; \backslash \backslash -16}{2}\] and \[x =\frac{-2 - 16}{2}\]Therefore, \[x = 7, -9\]

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 91影视!

Key Concepts

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

Derivative
The derivative of a function represents its rate of change. It's a fundamental concept in calculus. Calculating the derivative involves applying differentiation rules, such as the power rule and the sum rule.
Quadratic Formula
The quadratic formula is a solution to the quadratic equation of the form \(ax^{2}+bx+c=0\). The formula is written as: \[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\].
Where:
  • a, b, and c are coefficients from the quadratic equation.
  • The symbol '卤' means 'plus or minus,' indicating two potential solutions.
In our example, the derivative simplifies to a quadratic equation, which we solved using the quadratic formula step by step.
Function Analysis
Function analysis involves studying the behavior of a function. By analyzing its critical points, local minima, local maxima, and inflection points, we can understand the graphical shape and features of the function better.
In this exercise, finding the critical points where the derivative equals zero gives us valuable information about the function's turning points.
Differentiation
Differentiation is the process of finding the derivative of a function. It can be applied to complex functions, breaking them down into simpler parts and finding the derivatives of each.
In our exercise, we used basic differentiation rules to find the derivative of \(x^{3} + 3x^{2} - 189x + 1\), leading us to a quadratic equation we could solve for the critical points.

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

Let \(f(v)\) be the gas consumption (in liters/km) of a car going at velocity \(v\) (in km/hour). In other words, \(f(v)\) tells you how many liters of gas the car uses to go one kilometer if it is traveling at \(v\) kilometers per hour. In addition, suppose that \(f(80)=0.05\) and \(f^{\prime}(80)=\) \(0.0004 .\) a. Let \(g(v)\) be the distance the same car goes on one liter of gas at velocity \(v\). What is the relationship between \(f(v)\) and \(g(v)\) ? Hence find \(g(80)\) and \(g^{\prime}(80)\). b. Let \(h(v)\) be the gas consumption in liters per hour of a car going at velocity \(v\). In other words, \(h(v)\) tells you how many liters of gas the car uses in one hour if it is going at velocity \(v\). What is the algebraic relationship between \(h(v)\) and \(f(v)\) ? Hence find \(h(80)\) and \(h^{\prime}(80)\). c. How would you explain the practical meaning of these function and derivative values to a driver who knows no calculus? Include units on each of the function and derivative values you discuss in your response.

Let \(f(x)=\frac{1}{4} x^{3}+4\). a. Sketch a graph of \(y=f(x)\) and explain why \(f\) is an invertible function. b. Let \(g\) be the inverse of \(f\) and determine a formula for \(g\). c. Compute \(f^{\prime}(x), g^{\prime}(x), f^{\prime}(2),\) and \(g^{\prime}(6) .\) What is the special relationship between \(f^{\prime}(2)\) and \(g^{\prime}(6) ?\) Why?

Find the derivative of \(h(t)=t \tan t+\sin t\) \(h^{\prime}(t)=\) _________

Find the derivative of \(y=\sqrt{x}\left(x^{2}+4\right)\). \(\frac{d y}{d x}=\) __________

Find the derivative of the function \(f(t),\) below. \(f(t)=\ln \left(t^{2}+7\right)\) \(f^{\prime}(t)=\) _________
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation
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.