ALEKS 360 COL.ALG (18 WEEK) >ACCESS<
2nd Edition
ISBN: 9781264109029
Author: Miller
Publisher: INTER MCG
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Chapter 3.4, Problem 51PE
To determine
To calculate: The number of possible positive and negative real zeros of,
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Students have asked these similar questions
Suppose f and g are the piecewise-defined functions defined
here. For each combination of functions in Exercises 51–56,
(a) find its values at x = -1, x = 0, x = 1, x = 2, and x = 3,
(b) sketch its graph, and (c) write the combination as a
piecewise-defined function.
f(x) = {
(2x + 1, ifx 0
g(x) = {
-x, if x 2
8(4):
51. (f+g)(x)
52. 3f(x)
53. (gof)(x)
56. g(3x)
54. f(x) – 1
55. f(x – 1)
For Exercises 23–24, use the remainder theorem to determine
if the given number c is a zero of the polynomial.
23. f(x) = 3x + 13x + 2x + 52x – 40
a. c = 2
b. c =
24. f(x) = x* + 6x + 9x? + 24x + 20
а. с 3D —5
b. c = 2i
In Exercises 13-14, find the domain of each function.
13. f(x) 3 (х +2)(х — 2)
14. g(x)
(х + 2)(х — 2)
In Exercises 15–22, let
f(x) = x? – 3x + 8 and g(x) = -2x – 5.
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.Similar questions
- For questions 10 – 11, use the table to answer the questions. It is set up to multiply two polynomials. (show your work)arrow_forwardIn Exercises 33–38, express the function, f, in simplified form. Assume that x can be any real number. 33. f(x) = V36(x + 2)² 34. f(x) = V81(x – 2)2 35. f(x) = V32(x + 2)³ 36. f(x) = V48(x – 2)³ 37. f(x) = V3x² – 6x + 3 38. f(x) = V5x2 – 10x + 5 %3Darrow_forwardAccording to the fundamental theorem of algebra, how many zeros does the function f(x) = 4x3 – x2 – 2x + 1 have? %3D - 01 O2arrow_forward
- In Exercises 11–18, use the function f defined and graphed below toanswer the questions. (a) Does f (-1) exist?arrow_forwardDetermine which functions have two real number zeros by calculating the discriminant, b2 – 4ac. Check all that apply. O fx) = x² + 6x + 8 O g(x) = x² + 4x + 8 O h(x) = x2 – 12x + 32 O k(x) = x2 + 4x – 1 O p(x) = 5x2 + 5x + 4 O t(x) = x2 – 2x – 15 -arrow_forwardEXERCISE 6 (PSAT – 20151028 Section 4, Q9) (Calculator permitted) f(x) = x + 4(x – 3) Which of the following is an equivalent form of the expression above that displays the zeros of the function as constants or coefficients in the expression? A) f(x) = x² + 4x – 12 B) f(x) = (x – 6)(x + 2) C) f(x) = (x + 6)(x – 2) D) f(x) = (x +2)² – 16arrow_forward
- Use the function f(x) = -5x +3 to compute and expand and simplify the following. a. f(3) : b. f(x) + h = c. f(x + h) d. f(x + h) - f(x) 3Darrow_forwardComplete the table by evaluating the function h(x) = (2x – 5)² for the given values of x. h(x) = (2x – 5)² - 3 121 - 2 81 - 1 1 9 2 1 3 Answer Keyb h(x) = (2r – 5)² - 3 121 - 2 81 - 1 1 9 2 1 3 *arrow_forwardPls help ASAParrow_forward
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