Concept explainers
a.
To find: The possible rational zeroes of the function
a.
Answer to Problem 38E
Possible rational zeroes of the function are
Explanation of Solution
Given information:
Given function is
Formula used:
Rational root test: According to this test, the list of all possible rational zeroes of a polynomial is,
Possible zeroes
Consider the function as,
Here, leading coefficient is
Factors of constant term 4 are
Factors of leading coefficient 4are
Then possible rational zeroes are,
Therefore, possible rational zeroes of the function are
b.
To draw: The graph of the function
b.
Explanation of Solution
Given information:
Given function is
Possible rational zeroes of the function are
Consider the function as
Graph of the function can be drawn as,
From the graph of the function, it is clear that zeroes of the function are
c.
All the real zeroes of the function
c.
Answer to Problem 38E
Real zeroes of the function are
Explanation of Solution
Given information:
Given function
Possible rational zeroes of the function are
Consider the given function as function
From part (a), possible rational zeroes of the function are
From part (b), it is clear is clear that the curve do not touch x − axis at
Hence, real zeroes of the function are
Chapter 2 Solutions
EBK PRECALCULUS W/LIMITS
- Find the length of the following curve. 3 1 2 N x= 3 -y from y 6 to y=9arrow_forward3 4/3 3213 + 8 for 1 ≤x≤8. Find the length of the curve y=xarrow_forwardGiven that the outward flux of a vector field through the sphere of radius r centered at the origin is 5(1 cos(2r)) sin(r), and D is the value of the divergence of the vector field at the origin, the value of sin (2D) is -0.998 0.616 0.963 0.486 0.835 -0.070 -0.668 -0.129arrow_forward
- 10 The hypotenuse of a right triangle has one end at the origin and one end on the curve y = Express the area of the triangle as a function of x. A(x) =arrow_forwardIn Problems 17-26, solve the initial value problem. 17. dy = (1+ y²) tan x, y(0) = √√3arrow_forwardcould you explain this as well as disproving each wrong optionarrow_forward
- could you please show the computation of this by wiresarrow_forward4 Consider f(x) periodic function with period 2, coinciding with (x) = -x on the interval [,0) and being the null function on the interval [0,7). The Fourier series of f: (A) does not converge in quadratic norm to f(x) on [−π,π] (B) is pointwise convergent to f(x) for every x = R П (C) is in the form - 4 ∞ +Σ ak cos(kx) + bk sin(kx), ak ‡0, bk ‡0 k=1 (D) is in the form ak cos(kx) + bk sin(kx), ak 0, bk 0 k=1arrow_forwardSolve the equation.arrow_forward
- 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