/*! 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 27 (a) Suppose that the polynomial ... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Suppose that the polynomial function \(f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}\) has exactly \(k\) critical points and \(f^{\prime \prime}(x) \neq 0\) for all critical points \(x\). Show that \(n-k\) is odd. (b) For each \(n,\) show that if \(n-k\) is odd, then there is a polynomial function \(f\) of degree \(n\) with \(k\) critical points, at each of which \(f^{\prime \prime}\) is non-zero. (c) Suppose that the polynomial function \(f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}\) has \(k_{1}\) local maximum points and \(k_{2}\) local minimum points. Show that \(k_{2}=k_{1}+1\) if \(n\) is even, and \(k_{2}=k_{1}\) if \(n\) is odd. (d) Let \(n, k_{1}, k_{2}\) be three integers with \(k_{2}=k_{1}+1\) if \(n\) is even, and \(k_{2}=k_{1}\) if \(n\) is odd, and \(k_{1}+k_{2} < n .\) Show that there is a polynomial function \(f\) of degree \(n,\) with \(k_{1}\) local maximum points and \(k_{2}\) local minimum points. Hint: Pick \(a_{1} < a_{2} < \cdots< a_{k_{1}+k_{2}}\) and try \(f^{\prime}(x)=\prod_{i=1}^{k_{1}+k_{2}}\left(x-a_{i}\right) \cdot\left(1+x^{2}\right)^{l}\) for an appropriate number \(l.\)

Short Answer

Expert verified
n - k is odd; such polynomials exist; k2 = k1 + 1 if n is even, k2 = k1 if n is odd; such polynomials with given k1, k2, and n exist.

Step by step solution

01

Understand the First Derivative

Given the polynomial function \[f(x) = x^n + a_{n-1} x^{n-1} + \cdots + a_0\] the first derivative is \[f'(x) = n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \cdots + a_1.\] Critical points occur where \(f'(x) = 0\).
02

Analyze the Sign of the Second Derivative

The second derivative, which is \[f''(x) = n (n-1) x^{n-2} + (n-1)(n-2) a_{n-1} x^{n-3} + \cdots + 2 a_2,\] is non-zero at all critical points, indicating that these points are either local maxima or minima.
03

Apply Rolle’s Theorem

By Rolle's Theorem, every interval between critical points must contain at least one root of \(f'(x)\), and between each pair of roots of \(f'(x)\), there is one change in the sign of \(f''(x)\). Thus, for \(k\) critical points, the polynomial degree \(n\) provides \(n-k\) intervals, making \(n-k\) necessarily an odd number.
04

Construct a Polynomial for Part (b)

To show that there exists a polynomial with given \(n\) and \(k\), consider \[f'(x) = \prod_{i=1}^k (x - a_i) \cdot (1 + x^2)^l,\] where \(a_1 < a_2 < \cdots < a_k\) and \(l\) is chosen such that the total degree is \(n-1\). Integrating \(f'(x)\) gives a polynomial \(f(x)\) of degree \(n\) with \(k\) critical points.
05

Analyze Local Maxima and Minima for Part (c)

For a polynomial of degree \(n\), if \(f\) has \(k_1\) local maxima and \(k_2\) local minima, then if \(n\) is even, \(f\) alternates between peaks and valleys, so there must be one more minimum than maximum, \(k_2 = k_1 + 1\). If \(n\) is odd, the number of maxima and minima are balanced, \(k_2 = k_1\).
06

Construct a Polynomial for Part (d)

Given \(n\), \(k_1\), and \(k_2\) with \(k_1 + k_2 < n\), consider \[f'(x) = \prod_{i=1}^{k_1+k_2} (x - a_i) \cdot (1 + x^2)^l\] with \(a_1 < a_2 < \cdots < a_{k_1+k_2}\). By choosing \(l\) such that the total degree is \(n-1\), this generates a polynomial \(f(x)\) of degree \(n\) exhibiting \(k_1\) local maxima and \(k_2\) local minima.

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Key Concepts

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

polynomial function
A polynomial function is an expression made up of variables and coefficients, structured in the form of sums of terms. Each term consists of a variable raised to a non-negative integer power and multiplied by a coefficient. For example, a polynomial of degree n can be written as:
\[f(x) = x^n + a_{n-1} x^{n-1} + \text{...} + a_0 \].
Here, each power of x is associated with a coefficient (e.g., \(a_{n-1}\), \(a_0\)). The degree of the polynomial (n) is determined by the highest exponent on the variable x.
In these polynomials, we’re often interested in the points where the slope of the function, or its rate of change, is zero. This is where the first derivative comes into play.
first derivative
The first derivative of a polynomial function represents the slope or the rate at which the function's value changes. Mathematically, if \( f(x) \) is a polynomial, its first derivative \( f'(x) \) is found using differentiation rules for each term:
\[ f'(x) = n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \text{...} + a_1 \].
  • The first derivative is used to identify critical points, where the derivative equals zero \(f'(x) = 0\).
  • These points are significant because they represent possible local maxima, local minima, or points of inflection.
By finding the x-values where \( f'(x) = 0 \), we get the critical points of the polynomial function.
second derivative
The second derivative of a polynomial function provides information about the concavity of the function. It is found by differentiating the first derivative:
\[ f''(x) = n(n-1) x^{n-2} + (n-1)(n-2) a_{n-1} x^{n-3} + \text{...} + 2 a_2 \].
  • If the second derivative \( f''(x) \) is positive at a critical point, the function is concave up, indicating a local minimum.
  • If \( f''(x) \) is negative, the function is concave down, indicating a local maximum.
  • If \( f''(x) = 0 \), further analysis is needed to determine the nature of the concavity.
Therefore, the second derivative helps us confirm whether a critical point is a local maximum or a local minimum.
Rolle's Theorem
Rolle’s Theorem is a fundamental result in calculus that states:
If a function \( f \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one point \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

This theorem is particularly useful when analyzing polynomial functions because it guarantees the existence of a root for the first derivative between any two critical points. This suggests the polynomial's behavior and provides constraints that help us understand the relationship between the number of critical points and the polynomial's degree.
local maxima
A local maximum of a polynomial function is a point \( x \) where the function reaches a peak within a small neighborhood. Mathematically, \( f \) has a local maximum at \( x = c \) if:
  • \( f'(c) = 0 \) (the slope is zero, indicating a critical point)
  • \( f''(c) < 0 \) (the second derivative is negative, confirming concavity down)
Local maxima are important as they represent the highest value the function attains within a specific interval. They allow us to understand the function's behavior in real-world scenarios, such as optimizing a quantity or analyzing natural phenomena.
local minima
A local minimum of a polynomial function is a point \( x \) where the function reaches the lowest value within a small neighborhood. Mathematically, \( f \) has a local minimum at \( x = c \) if:
  • \( f'(c) = 0 \)
  • \( f''(c) > 0 \) (indicating concavity up)
Local minima are just as crucial as local maxima for understanding the function's behavior. They help in minimizing costs, reducing risks, or understanding valleys in natural terrains. Tracking local minima and maxima provides a full picture of the polynomial's critical behavior over its domain.

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Most popular questions from this chapter

(a) A point \(x\) is called a strict maximum point for \(f\) on \(A\) if \(f(x)>f(y)\) for all \(y\) in \(A\) with \(y \neq x\) (compare with the definition of an ordinary maximum point). A local strict maximum point is defined in the obvious way. Find all local strict maximum points of the function $$f(x)=\left\\{\begin{array}{ll} 0, & x \text { irrational } \\ \frac{1}{q}, & x=\frac{p}{q} \text { in lowest terms }. \end{array}\right.$$ It seems quite unlikely that a function can have a local strict maximum at every point (although the above example might give one pause for thought). Prove this as follows. (b) Suppose that every point is a local strict maximum point for \(f\). Let \(x_{1}\) be any number and choose \(a_{1} < x_{1} < b_{1}\) with \(b_{1}-a_{1}<1\) such that \(f\left(x_{1}\right)>f(x)\) for all \(x\) in \(\left[a_{1}, b_{1}\right] .\) Let \(x_{2} \neq x_{1}\) be any point in \(\left(a_{1}, b_{1}\right)\) and choose \(a_{1} \leq a_{2}< x_{2} < b_{2} \leq b_{1}\) with \(b_{2}-a_{2} < \frac{1}{2}\) such that \(f\left(x_{2}\right)>f(x)\) for all \(x\) in \(\left[a_{2}, b_{2}\right] .\) Continue in this way, and use the Nested Interval Theorem (Problem 8-14 ) to obtain a contradiction.

(a) Prove that two polynomial functions of degree \(m\) and \(n\), respectively, intersect in at most \(\max (m, n)\) points. (b) For each \(m\) and \(n\) exhibit two polynomial functions of degree \(m\) and \(n\) which intersect \(\max (m, n)\) times.

It is easy to find a function \(f\) such that \(|f|\) is differentiable but \(f\) is not. For example, we can choose \(f(x)=1\) for \(x\) rational and \(f(x)=-1\) for \(x\) irrational. In this example \(f\) is not even continuous, nor is this a mere coincidence: Prove that if \(|f|\) is differentiable at \(a,\) and \(f\) is continuous at \(a\) then \(f\) is also differentiable at \(a\). Hint: It suffices to consider only \(a\) with \(f(a)=0 .\) Why? In this case, what must \(|f|^{\prime}(a)\) be?

A function \(f\) is increasing at \(a\) if there is some number \(\delta>0\) such $$f(x)>f(a) \text { if } a< x< a+\delta$$ and $$f(x)< f(a) \text { if } a-\delta < x < a$$ Notice that this does not mean that \(f\) is increasing in the interval \((a-\delta .\) \(a+\delta) ;\) for example, the function shown in Figure 34 is increasing at \(0,\) but is not an increasing function in any open interval containing 0. (a) Suppose that \(f\) is continuous on [0.1] and that \(f\) is increasing at \(a\) for every \(a\) in \([0.1] .\) Prove that \(f\) is increasing on \([0.1] .\) (First convince yourself that there is something to be proved.) Hint: For \(0< b <1\) prove that the minimum of \(f\) on \([b, 1]\) must be at \(b\). (b) Prove part (a) without the assumption that \(f\) is continuous, by considering for each \(b\) in [0.1] the set \(S_{b}=\\{x: f(y) \geq f(b) \text { for all } y \text { in }[b, x]\\}\) (This part of the problem is not necessary for the other parts.) Hint: Prove that \(S_{b}=\\{x: b \leq x \leq 1\\}\) by considering sup \(S_{b}.\) (c) If \(f\) is increasing at \(a\) and \(f\) is differentiable at \(a\), prove that \(f^{\prime}(a) \geq 0\) (this is easy). (d) II \(f^{\prime}(a)>0,\) prove that \(f\) is increasing at \(a\) (go right back to the definition of \(f^{\prime}(a)\).) (e) Use parts (a) and (d) to show, without using the Nean Value Theorem, that if \(f\) is continuous on \([0,1]\) and \(f^{\prime}(a)>0\) for all \(a\) in \([0.1],\) then \(f\) is increasing on \([0,1].\) (f) Suppose that \(f\) is continuous on [0,1] and \(f^{\prime}(a)=0\) for all \(a\) in (0,1) Apply part (e) to the function \(g(x)=f(x)+\varepsilon x\) to show that \(f(1)-\) \(f(0)>-\varepsilon .\) Similarly, show that \(f(1)-f(0)<\varepsilon\) by considering \(h(x)=\varepsilon x-f(x) .\) Conclude that \(f(0)=f(1).\) This particular proof that a function with zero derivative must be constant has many points in common with a proof of H. A. Schwarz, which may be the first rigorous proof ever given. Its discoverer, at least, seemed to think it was. See his exubcrant letter in reference [54] of the Suggested Reading.

Find the following limits: (i) \(\lim _{x \rightarrow 0} \frac{x}{\tan x}.\) (ii) \(\lim _{x \rightarrow 0} \frac{\cos ^{2} x-1}{x^{2}}.\)

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