/*! 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 21 Find the inflection points of \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the inflection points of \(f(x)=x^{4}+x^{3}-3 x^{2}+2\).

Short Answer

Expert verified
The inflection points are at \(x = -1\) and \(x = \frac{1}{2}\).

Step by step solution

01

Find the second derivative

Inflection points occur where the second derivative of the function changes sign. Start by differentiating the given function twice. The original function is:\[ f(x) = x^4 + x^3 - 3x^2 + 2 \]First, find the first derivative, \(f'(x)\):\[ f'(x) = 4x^3 + 3x^2 - 6x \]Then, find the second derivative, \(f''(x)\):\[ f''(x) = 12x^2 + 6x - 6 \]
02

Set the second derivative to zero

To find potential inflection points, set the second derivative equal to zero and solve for \(x\):\[ 12x^2 + 6x - 6 = 0 \]Simplify by dividing the entire equation by 6:\[ 2x^2 + x - 1 = 0 \]
03

Solve the quadratic equation

Solve the quadratic equation \(2x^2 + x - 1 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = 1\), and \(c = -1\).Plug in the values:\[x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}\]This simplifies to:\[x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4} = \frac{-1 \pm 3}{4}\]Thus, the solutions are:\[ x = \frac{2}{4} = \frac{1}{2} \] and \[ x = \frac{-4}{4} = -1 \]
04

Check second derivative sign change

Verify a change in sign of \(f''(x)\) at each solution point to confirm they are inflection points.For \(x = \frac{1}{2}\) and \(x = -1\), check intervals:1. For \(x < -1\), choose \(x = -2\), then \(f''(-2) = 36 - 12 - 6 = 18 > 0\) (positive).2. For \(x = -1\), \(f''(-1) = 12(-1)^2 + 6(-1) - 6 = 12 - 6 - 6 = 0\).3. For \(-1 < x < \frac{1}{2}\), choose \(x = 0\), then \(f''(0) = -6\) (negative).4. For \(x = \frac{1}{2}\), \(f''(\frac{1}{2}) = 12(\frac{1}{2})^2 + 6(\frac{1}{2}) - 6 = 3 + 3 - 6 = 0\).5. For \(x > \frac{1}{2}\), choose \(x = 1\), then \(f''(1) = 12 + 6 - 6 = 12 > 0\) (positive).Both \(x = -1\) and \(x = \frac{1}{2}\) see a sign change in the second derivative, confirming them as inflection points.

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Key Concepts

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

Second Derivative
The second derivative of a function helps to uncover valuable information about the shape and behavior of a graph. Specifically, it tells us about the curvature, or concavity, of a function.
- If the second derivative, denoted as \(f''(x)\), is positive at a certain interval, the graph is concave up, resembling a bowl facing upwards.- Conversely, if \(f''(x)\) is negative, the graph is concave down, like an upside-down bowl.
Inflection points occur where \(f''(x)\) changes sign. These points are crucial because they indicate where the graph switches from concave up to concave down, or vice versa. Finding an inflection point involves calculating the second derivative and determining where it equals zero or does not exist. For our exercise, the function was differentiated twice to find \(f''(x)=12x^2 + 6x - 6\). Identifying where this result changes sign reveals potential inflection points.
Quadratic Formula
A fundamental tool in solving quadratic equations is the quadratic formula. Given a quadratic equation of the form \(ax^2 + bx + c = 0\), the solutions for \(x\) can be found using:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula pulls roots directly from the coefficients \(a\), \(b\), and \(c\). It is very effective, especially when factoring is not straightforward.
In our exercise, the second derivative, \(2x^2 + x - 1 = 0\), was solved using this formula. Plugging into the formula, we arrive at the solutions, illustrating that this method is a powerful shortcut over manual factoring. Understanding how to use the quadratic formula opens doors to solving a variety of mathematical problems easily.
Sign Change
The concept of a sign change plays a pivotal role in determining the inflection points. Inflection points indicate where a graph changes its concavity, and this aspect relates directly to the sign of the second derivative.
- A change from positive to negative indicates the graph transitions from concave up to concave down.- A switch from negative to positive reveals a shift from being concave down to concave up.
In our specific problem, the computation showed a change in the sign of the second derivative at \(x = -1\) and \(x = \frac{1}{2}\). At these points, the sign of \(f''(x)\) shifts, confirming them as inflection points. These checks clarify important aspects of the graph's behavior and are essential for fully grasping the curve's properties.
Calculus
Calculus provides powerful techniques for understanding and analyzing functions. At its core lie derivatives, which describe how functions change. - The first derivative \(f'(x)\), informs us about the slope of the function or how steep it is at any given point.- The second derivative \(f''(x)\) takes this further by detailing the curvature, telling you if it is bending upwards or downwards.
The process of differentiating both first and second derivatives reveals vital information about a function's graph, including maxima, minima, and inflection points.
In this exercise, calculus guided us step-by-step to locate inflection points by differentiating twice to observe changes in signs of \(f''(x)\). This approach not only highlights the elegance of calculus but its utility in real-world applications, offering insights into trends and behaviors within data and mathematical models alike.

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Most popular questions from this chapter

An ice cream company finds that at a price of \(\$ 4.00\), demand is 4000 units. For every \(\$ 0.25\) decrease in price, demand increases by 200 units. Find the price and quantity sold that maximize revenue.

The elasticity of a good is \(E=0.5 .\) What is the effect on the quantity demanded of: (a) A \(3 \%\) price increase? (b) A \(3 \%\) price decrease?

A company can produce and sell \(f(L)\) tons of a product per month using \(L\) hours of labor per month. The wage of the workers is \(w\) dollars per hour, and the finished product sells for \(p\) dollars per ton. (a) The function \(f(L)\) is the company's production function. Give the units of \(f(L) .\) What is the practical significance of \(f(1000)=400 ?\) (b) The derivative \(f^{\prime}(L)\) is the company's marginal product of labor. Give the units of \(f^{\prime}(L) .\) What is the practical significance of \(f^{\prime}(1000)=2 ?\) (c) The real wage of the workers is the quantity of product that can be bought with one hour's wages. Show that the real wage is \(w / p\) tons per hour. (d) Show that the monthly profit of the company is $$ \pi(L)=p f(L)-w L $$ (e) Show that when operating at maximum profit, the company's marginal product of labor equals the real wage: $$ f^{\prime}(L)=\frac{w}{p} $$

Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. \(f(x)=3 x^{5}-5 x^{3}\)

(a) On a computer or calculator, graph \(f(\theta)=\theta-\sin \theta\). Can you tell whether the function has any zeros in the interval \(0 \leq \theta \leq 1 ?\) (b) Find \(f^{\prime}\). What does the sign of \(f^{\prime}\) tell you about the zeros of \(f\) in the interval \(0 \leq \theta \leq 1 ?\)

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