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Chapter 3: Problem 2

Find the inflection points of \(f(x)=2 x^{4}+27 x^{3}-21 x^{2}+15\). (Give your answers as a comma separated list, e.g., \(3,-2 .)\) inflection points = ____ .

Short Answer

Expert verified
The inflection points are at \( \frac{1}{4}, -7 \).

Step by step solution

01

- Find the second derivative

To find the inflection points, first find the second derivative of the function. Start by finding the first derivative of the given function: \[ f(x) = 2x^4 + 27x^3 - 21x^2 + 15 \] The first derivative is: \[ f'(x) = 8x^3 + 81x^2 - 42x \] Now, find the derivative of \( f'(x) \) to get the second derivative: \[ f''(x) = 24x^2 + 162x - 42 \]
02

- Set the second derivative to zero

To find potential inflection points, set the second derivative equal to zero and solve for \( x \): \[ 24x^2 + 162x - 42 = 0 \]
03

- Solve the quadratic equation

Solve the quadratic equation \( 24x^2 + 162x - 42 = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 24 \), \( b = 162 \), and \( c = -42 \). \[ x = \frac{-162 \pm \sqrt{162^2 - 4 \cdot 24 \cdot (-42)}}{2 \cdot 24} \] Calculate the discriminant first: \[ b^2 - 4ac = 162^2 - 4 \cdot 24 \cdot (-42) = 26244 + 4032 = 30276 \] So, the solution for \( x \) is: \[ x = \frac{-162 \pm \sqrt{30276}}{48} \] \[ x = \frac{-162 \pm 174}{48} \] \[ x = \frac{12}{48} \quad \text{or} \quad x = \frac{-336}{48} \] \[ x = \frac{1}{4} \quad \text{or} \quad x = -7 \]
04

- Verify the concavity changes

To confirm that these points are inflection points, check the concavity on either side of \( x = \frac{1}{4} \) and \( x = -7 \) by picking test points in the intervals defined by these x-values. Determine if the second derivative changes sign (indicating a change in concavity).

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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 gives us information about the concavity of that function. Finding the second derivative is a process that starts with computing the first derivative.

For example, let's take the function: \[ f(x) = 2x^4 + 27x^3 - 21x^2 + 15 \]
First, we find the first derivative: \[ f'(x) = 8x^3 + 81x^2 - 42x \]
Next, we find the second derivative by differentiating the first derivative: \[ f''(x) = 24x^2 + 162x - 42 \]
With the second derivative expression in hand, we can analyze the concavity and locate inflection points by solving \[ f''(x) = 0 \].
concavity
Concavity explains how the curve bends. A function can be 'concave up' or 'concave down'.

When a function is concave up, it looks like a U (shaped upwards). When concave down, it resembles an upside-down U.

The second derivative helps determine this:
  • If \[ f''(x) > 0 \], the function is concave up.

  • If \[ f''(x) < 0 \], the function is concave down.

For example, with \[ f''(x) = 24x^2 + 162x - 42 \], to find where the concavity changes, set \[ f''(x) = 0 \] and solve for \[ x \]. The solutions are potential inflection points where the concavity might change.
quadratic formula
To solve equations like \[ 24x^2 + 162x - 42 = 0 \], we use the quadratic formula. This is very useful for second-degree polynomials.

The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here is how it is used for our second derivative equation:
  • Identify coefficients: \[ a = 24 \], \[ b = 162 \], and \[ c = -42 \].

  • Calculate discriminant: \[ b^2 - 4ac = 26244 + 4032 = 30276 \].

  • Solve for \[ x \]: \[ x = \frac{-162 \pm \sqrt{30276}}{48} \] results in \[ x = \frac{1}{4} \] or \[ x = -7 \].

These solutions are our potential inflection points.
calculus problem solving
Solving calculus problems requires a methodical approach. Here's how to tackle problems like finding inflection points:

  1. Differentiate: Compute the first and then the second derivative.

  2. Set to Zero: Set the second derivative equal to zero to find potential inflection points.

  3. Solve: Use methods like the quadratic formula to solve for x.

  4. Verify: Ensure that there is a change in concavity by checking the second derivative's sign on either side of each solution.

In our exercise, we found potential inflection points at \[ x = \frac{1}{4} \] and \[ x = -7 \]. We confirm these by checking concavity. If the concavity changes, these are indeed inflection points.

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Consider the one-parameter family of functions given by \(p(x)=x^{3}-a x^{2},\) where \(a>0\) a. Sketch a plot of a typical member of the family, using the fact that each is a cubic polynomial with a repeated zero at \(x=0\) and another zero at \(x=a\). b. Find all critical numbers of \(p\). c. Compute \(p^{\prime \prime}\) and find all values for which \(p^{\prime \prime}(x)=0 .\) Hence construct a second derivative sign chart for \(p\). d. Describe how the location of the critical numbers and the inflection point of \(p\) change as \(a\) changes. That is, if the value of \(a\) is increased, what happens to the critical numbers and inflection point?
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