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Chapter 3: Problem 8

In Exercises \(7-12,\) find the horizontal tangents of the curve. $$y=x^{3}-4 x^{2}+x+2$$

Short Answer

Expert verified
The coordinates for the horizontal tangents of the curve are \(\left(\frac{8 + \sqrt{52}}{6}, \left(\frac{8 + \sqrt{52}}{6}\right)^3 - 4*\left(\frac{8 + \sqrt{52}}{6}\right)^2 + \left(\frac{8 + \sqrt{52}}{6}\right) + 2\right)\) and \(\left(\frac{8 - \sqrt{52}}{6}, \left(\frac{8 - \sqrt{52}}{6}\right)^3 - 4*\left(\frac{8 - \sqrt{52}}{6}\right)^2 + \left(\frac{8 - \sqrt{52}}{6}\right) + 2\right)\)

Step by step solution

01

Taking Derivative

First, take the derivative of \(y = x^{3}-4 x^{2}+x+2\). The derivative, denoted as \(y'\), is obtained by applying the power rule, which states that d/dx[ \(x^n\) ] = \(n*x^{n-1}\). Hence, \(y' = 3x^{2}-8x+1\).
02

Finding the Potential Points for Horizontal Tangents

To find the horizontal tangents, set the derivative \(y'\) equal to zero and solve for \(x\), because the points where the derivative is zero correspond to potential horizontal tangents. So the equation is: \(3x^{2}-8x+1=0\). This is a quadratic equation and it's solved by using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, a = 3, b = -8 and c = 1.
03

Solve for x

Apply the quadratic formula to solve for \(x\). \(x = \frac{8 \pm \sqrt{(-8)^{2} - 4*3*1}}{2*3}\). Calculating under the square root gives \(x = \frac{8 \pm \sqrt{64 - 12}}{6}\) which simplifies to \(x = \frac{8 \pm \sqrt{52}}{6}\). Two solutions for \(x\) are obtained giving \(x = \frac{8 + \sqrt{52}}{6}\) and \(x = \frac{8 - \sqrt{52}}{6}\).
04

Find the y-coordinates for the points

Substitute the x-values obtained from step 3 back into the original equation \(y = x^{3}-4 x^{2}+x+2\) to find the corresponding y-values. Thus two points of horizontal tangent are obtained.

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Key Concepts

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

Calculus
Calculus, often when first encountered, might look daunting, but it's just a branch of mathematics that deals with changes. In essence, it's the study of how things vary, either in real-world situations or theoretical scenarios. It comprises two main parts: differentiation and integration. Differentiation, as used in the exercise, focuses on how a function changes at a certain point.
  • Differentiation: This process helps find the rate at which something changes. In the example of finding tangent lines, differentiation helps determine the slope of the curve at any point.
  • Integration: Although not used in this exercise, integration is about accumulation and finding total values, such as areas under curves.
Calculus is essential in solving complex problems across various scientific and engineering fields, allowing us to make predictions and understand dynamics fully.
Tangent Lines
The concept of tangent lines ties closely to differentiation. A tangent line touches a curve at a single point without crossing it. Imagine standing on a hill; if you draw a line directly under your feet on the hill's slope, that's your tangent line. It represents the slope or steepness of the curve at that point.
  • Horizontal Tangents: These are special because their slope is zero. A horizontal tangent means the curve does not rise or fall at that point, indicating a peak, valley, or level section.
  • Finding Tangents: To find a tangent line's equation, you usually start by taking the derivative, which gives you a formula for the slope at any point on the curve.
In the exercise, locating horizontal tangents involves setting the derivative equal to zero and solving for the points where this occurs.
Quadratic Equations
Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). They are incredibly common and can describe a wide variety of problems, from the path of a thrown ball to optimization challenges in business. In our exercise, the quadratic equation appears when finding the horizontal tangents on the curve.
  • General Form: The basic form is \( ax^2 + bx + c = 0 \), making them easy to recognize.
  • Quadratic Formula: It's a reliable method to find solutions for quadratic equations: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula helps us find the points where the derivative equals zero, leading to the specific coordinates of interest on the curve.
Quadratics provide solutions to many practical problems and are a vital component of algebra and calculus applications.

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Most popular questions from this chapter

Draining a Tank It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth y of fluid in the tank t hours after the valve is opened is given by the formula \(y=6\left(1-\frac{t}{12}\right)^{2} \mathrm{m}\) (a) Find the rate \(d y / d t(\mathrm{m} / \mathrm{h})\) at which the water level is changing at time. (b) When is the fluid level in the tank falling fastest? slowest? What are the values of \(d y / d t\) at these times? (c) Graph \(y\) and \(d y / d t\) together and discuss the behavior of \(y\) in relation to the signs and values of \(d y / d t .\)

Find the normals to the curve \(x y+2 x-y=0\) that are parallel to the line $2 x+y=0 .

In Exercises \(37-42,\) find \(f^{\prime}(x)\) and state the domain of \(f^{\prime}\) $$f(x)=\log _{2}(3 x+1)$$

Find A\( and \)B\( in \)y=A \sin x+B \cos x\( so that \)y^{\prime \prime}-y=\sin x

In Exercises \(33-36,\) find \(d y / d x\) $$y=x^{1+\sqrt{2}}(1+\sqrt{2}) x^{\sqrt{2}}$$
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