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Chapter 3: Problem 16

Find the relative extrema of the function, if they exist. ist your answers in terms of ordered pairs. Then sketch a graph of the function. $$ f(x)=x^{4}-8 x^{2}+3 $$

Short Answer

Expert verified
(0, 3) is a relative maximum. (2, -13) and (-2, -13) are relative minima.

Step by step solution

01

- Find the first derivative

To determine the relative extrema, start by finding the first derivative of the function. For the function \(f(x) = x^4 - 8x^2 + 3\), the first derivative is assessed as follows:\[f'(x) = \frac{d}{dx}(x^4 - 8x^2 + 3)\]Using the power rule, we get:\[f'(x) = 4x^3 - 16x\]
02

- Set the first derivative to zero

To find critical points, set the first derivative equal to zero and solve for x:\[4x^3 - 16x = 0\]Factor the equation:\[4x(x^2 - 4) = 0\]This gives us:\[4x(x - 2)(x + 2) = 0\]Therefore, the solutions are:\[x = 0, x = 2, x = -2\]
03

- Find the second derivative

Next, we need the second derivative to apply the second derivative test. The second derivative of \(f(x)\) is:\[f''(x) = \frac{d}{dx}(4x^3 - 16x)\]Taking the derivative again using the power rule, we get:\[f''(x) = 12x^2 - 16\]
04

- Apply the second derivative test

Evaluate the second derivative at each critical point. If \(f''(x) > 0\), then \(f\) has a relative minimum at that point. If \(f''(x) < 0\), then \(f\) has a relative maximum. If \(f''(x) = 0\), the test is inconclusive.For \(x = 0:\)\[f''(0) = 12(0)^2 - 16 = -16 < 0\]So, there is a relative maximum at \(x = 0\).For \(x = 2:\)\[f''(2) = 12(2)^2 - 16 = 32 > 0\]So, there is a relative minimum at \(x = 2\).For \(x = -2:\)\[f''(-2) = 12(-2)^2 - 16 = 32 > 0\]So, there is a relative minimum at \(x = -2\).
05

- Find the y-values for the critical points

To get the coordinates of the relative extrema, substitute the x-values back into the original function \(f(x)\):For \(x = 0:\)\[f(0) = (0)^4 - 8(0)^2 + 3 = 3\]So the relative maximum is \((0, 3)\).For \(x = 2:\)\[f(2) = (2)^4 - 8(2)^2 + 3 = 16 - 32 + 3 = -13\]So the relative minimum is \((2, -13)\).For \(x = -2:\)\[f(-2) = (-2)^4 - 8(-2)^2 + 3 = 16 - 32 + 3 = -13\]So the relative minimum is \((-2, -13)\).
06

- Sketch the graph

Use the relative extrema points found to sketch the graph of the function. Plot the points \((0, 3)\), \((2, -13)\), and \((-2, -13)\). Make sure to show that the graph has a relative maximum at \((0, 3)\) and relative minima at \((2, -13)\) and \((-2, -13)\).

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Key Concepts

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

Critical Points
Critical points are pivotal in finding the relative extrema of a function. They occur where the first derivative of the function is zero or undefined. For a given function, say \(f(x) = x^4 - 8x^2 + 3\), you'll begin by finding its first derivative \(f'(x)\). Using the power rule, this derivative is \(f'(x) = 4x^3 - 16x\). Next, set \(f'(x) = 0\) to find the x-values where the function's slope is zero. For our function:
  • \(4x(x^2 - 4) = 0\)
  • \(x = 0, x = 2, x = -2\)
These x-values are the critical points and potential candidates for relative extrema.
First Derivative Test
The first derivative test helps determine whether each critical point is a relative minimum, maximum, or neither. After you find the critical points, you investigate the intervals around these points to see how the sign of the first derivative changes:
  • If \(f'(x)\) changes from positive to negative, \(f(x)\) has a relative maximum at that point.
  • If \(f'(x)\) changes from negative to positive, \(f(x)\) has a relative minimum at that point.
  • If \(f'(x)\) does not change sign, the critical point is neither a maximum nor a minimum.
While the first derivative test gives valid results, it can be more cumbersome for complex functions. An alternative and often quicker method is the second derivative test.
Second Derivative Test
The second derivative test offers a more straightforward validation of the nature of critical points. Once the critical points are determined, compute the second derivative of the function. For our example:
  • The first derivative is \(f'(x) = 4x^3 - 16x\).
  • The second derivative is \(f''(x) = 12x^2 - 16\).
Now, substitute each critical point into \(f''(x)\):
  • For \(x = 0\): \(f''(0) = 12(0)^2 - 16 = -16 < 0\), indicating a relative maximum.
  • For \(x = 2\): \(f''(2) = 12(2)^2 - 16 = 32 > 0\), indicating a relative minimum.
  • For \(x = -2\): \(f''(-2) = 12(-2)^2 - 16 = 32 > 0\), also indicating a relative minimum.
Polynomial Functions
Understanding polynomial functions makes the process of finding extrema more accessible. A polynomial function is of the form \(f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0\). Key characteristics include:
  • Continuity and smoothness: Polynomials are continuous and differentiable over their domain.
  • Degrees: The degree of the polynomial affects its shape and the number of extrema it can have.
  • Symmetry: Even-degree polynomials may exhibit symmetry about the y-axis, while odd-degree polynomials may show rotational symmetry about the origin.
For example, \(f(x) = x^4 - 8x^2 + 3\) is a fourth-degree polynomial. Its general shape allows for at most three turning points and relative extrema, determined by differentiating and solving the critical points.

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When we speak, the position of the tongue changes the shapes of the mouth and throat cavity, thereby changing the sound that is generated. For example, to pronounce the first vowel of the word father, the mouth is much wider than the throat cavity. For many vowel sounds, the vocal tract may be modeled by two adjoining tubes, with one end closed (the glottis) and the other end open (the lips). We denote by \(A_{m}\) and \(L_{m}\) the cross-sectional area and length of the mouth, and we denote by \(A_{t}\) and \(L_{t}\) the cross-sectional area and length of the throat cavity. The shape of the vocal tract tends to promote certain sound frequencies. \({ }^{15}\) Let \(c\) be the speed of sound. If \(\int\) is a frequency promoted by the vocal tract (in Hertz) and we let \(x=2 \pi \int / c\), then \(x\) is a solution of $$ \frac{1}{A_{m}} \tan \left(L_{m} x\right)-\frac{1}{A_{t}} \cot \left(L_{t} x\right)=0 $$ a) For a certain speaker of the first vowel of the word father; \(A_{m}=10 A_{t}\) (that is, the mouth opening is 10 times larger than the throat opening),\(L_{m}=8 \mathrm{~cm}\), and \(L_{t}=9.7 \mathrm{~cm} .^{16}\) Use this information to show that $$ \tan (8 x)-10 \cot (9.7 x)=0 $$ a) For a certain speaker of the first vowel of the word father; \(A_{m}=10 A_{t}\) (that is, the mouth opening is 10 times larger than the throat opening),\(L_{m}=8 \mathrm{~cm}\), and \(L_{t}=9.7 \mathrm{~cm} .^{16}\) Use this information to show that $$ \tan (8 x)-10 \cot (9.7 x)=0 $$ b) Use a grapher to sketch the graph of \(y=\tan (8 x)-10 \cot (9.7 x)\) using the graphing window \([0,0.5,-2,2] .\) From the graph, estimate the first three \(x\) -intercepts. c) Use Newton's method to find the first three solutions. Use \(x_{1}=0.1, x_{1}=0.2\), and \(x_{1}=0.45\) as your three starting points. d) The speed of sound is approximately \(35,400 \mathrm{~cm} / \mathrm{s}\). Use this value for \(c\) and the relationship \(x=2 \pi f / c\) to find the first three natural frequencies of the speaker's vocal tract.
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