Concept explainers
(a)
The real zeros of the polynomial function.
(a)
Answer to Problem 39E
The real zero of the polynomial function is at x = -2 and x = 1
Explanation of Solution
Given information:
The given function
Formula used:
The real zeroes of the polynomial function by putting
Calculation:
It can be shown that for a polynomial function f of degree n, the following statements are true.
(1) The function has at most n real zeroes
(2) The graph of f has, at most n-1 turning points.
We can determine the real zeroes of the polynomial function by putting
So, the real zeroes for the given function are x = -2 and x =1
Hence, the real zero of the polynomial function is at x = -2 and x = 1
Conclusion:
The real zero of the polynomial function is at x = -2 and x = 1
(b)
The multiplicity of each zero is even or odd.
(b)
Answer to Problem 39E
The multiplicity is odd and there is only one turning point
Explanation of Solution
Given information:
The given function
Formula used:
The multiplicity of each zero is odd
Calculation:
It can be shown that for a polynomial function f of degree n, the following statements are true.
(1) The function has at most n real zeroes
(2) The graph of f has, at most n-1 turning points.
The degree of the given polynomial function is 2. So, there will be at most one turning point and multiplicity of each zero is odd.
Hence, the multiplicity is odd and there is only one turning point
Conclusion:
The multiplicity is odd and there is only one turning point
(c)
The maximum possible number of turning points of the graph of the function.
(c)
Answer to Problem 39E
Explanation of Solution
Given information:
The given function
Formula used:
The graph is plotted against the x axis and y axis.
Calculation:
It can be shown that for a polynomial function f of degree n, the following statements are true.
(1) The function has at most n real zeroes
(2) The graph of f has, at most n-1 turning points.
Let us draw the graph of the given polynomial function,
We can observe from the graph that the zeroes of the polynomial function are at x = -2 and
x = 1 and there is only one turning point.
Conclusion:
The zeroes of the polynomial function are at x = -2 and x = 1 and there is only one turning point.
Chapter 2 Solutions
EBK PRECALCULUS W/LIMITS
- nd ave a ction and ave an 48. The domain of f y=f'(x) x 1 2 (= x<0 x<0 = f(x) possible. Group Activity In Exercises 49 and 50, do the following. (a) Find the absolute extrema of f and where they occur. (b) Find any points of inflection. (c) Sketch a possible graph of f. 49. f is continuous on [0,3] and satisfies the following. X 0 1 2 3 f 0 2 0 -2 f' 3 0 does not exist -3 f" 0 -1 does not exist 0 ve tes where X 0 < x <1 1< x <2 2arrow_forwardNumerically estimate the value of limx→2+x3−83x−9, rounded correctly to one decimal place. In the provided table below, you must enter your answers rounded exactly to the correct number of decimals, based on the Numerical Conventions for MATH1044 (see lecture notes 1.3 Actions page 3). If there are more rows provided in the table than you need, enter NA for those output values in the table that should not be used. x→2+ x3−83x−9 2.1 2.01 2.001 2.0001 2.00001 2.000001arrow_forwardFind the general solution of the given differential equation. (1+x)dy/dx - xy = x +x2arrow_forwardEstimate the instantaneous rate of change of the function f(x) = 2x² - 3x − 4 at x = -2 using the average rate of change over successively smaller intervals.arrow_forwardGiven the graph of f(x) below. Determine the average rate of change of f(x) from x = 1 to x = 6. Give your answer as a simplified fraction if necessary. For example, if you found that msec = 1, you would enter 1. 3' −2] 3 -5 -6 2 3 4 5 6 7 Ꮖarrow_forwardGiven the graph of f(x) below. Determine the average rate of change of f(x) from x = -2 to x = 2. Give your answer as a simplified fraction if necessary. For example, if you found that msec = , you would enter 3 2 2 3 X 23arrow_forwardA function is defined on the interval (-π/2,π/2) by this multipart rule: if -π/2 < x < 0 f(x) = a if x=0 31-tan x +31-cot x if 0 < x < π/2 Here, a and b are constants. Find a and b so that the function f(x) is continuous at x=0. a= b= 3arrow_forwardUse the definition of continuity and the properties of limits to show that the function is continuous at the given number a. f(x) = (x + 4x4) 5, a = -1 lim f(x) X--1 = lim x+4x X--1 lim X-1 4 x+4x 5 ))" 5 )) by the power law by the sum law lim (x) + lim X--1 4 4x X-1 -(0,00+( Find f(-1). f(-1)=243 lim (x) + -1 +4 35 4 ([ ) lim (x4) 5 x-1 Thus, by the definition of continuity, f is continuous at a = -1. by the multiple constant law by the direct substitution propertyarrow_forward1. Compute Lo F⚫dr, where and C is defined by F(x, y) = (x² + y)i + (y − x)j r(t) = (12t)i + (1 − 4t + 4t²)j from the point (1, 1) to the origin.arrow_forward2. Consider the vector force: F(x, y, z) = 2xye²i + (x²e² + y)j + (x²ye² — z)k. (A) [80%] Show that F satisfies the conditions for a conservative vector field, and find a potential function (x, y, z) for F. Remark: To find o, you must use the method explained in the lecture. (B) [20%] Use the Fundamental Theorem for Line Integrals to compute the work done by F on an object moves along any path from (0,1,2) to (2, 1, -8).arrow_forwardhelp pleasearrow_forwardIn each of Problems 1 through 4, draw a direction field for the given differential equation. Based on the direction field, determine the behavior of y as t → ∞. If this behavior depends on the initial value of y at t = 0, describe the dependency.1. y′ = 3 − 2yarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning