Problem 1: (a) Show that Let f(x) = x³ + 5x² - 4x - 20 5 = 5 4 91(x) = x³ + x² 92(x) = √√ (20 +4x − x³) - both have fixed points p when f(p) = 0 (i.e. a fixed point that is a root of f). (b) Find a value of g₁(x) that does not lie in [1,3]. Hence conclude that the conditions of the Fixed Point Theorem are not satisfied. (c) However, by plotting y = 9₁(x) and y = x on the same axis, observe that g₁ has a unique fixed point in [1,3].
Problem 1: (a) Show that Let f(x) = x³ + 5x² - 4x - 20 5 = 5 4 91(x) = x³ + x² 92(x) = √√ (20 +4x − x³) - both have fixed points p when f(p) = 0 (i.e. a fixed point that is a root of f). (b) Find a value of g₁(x) that does not lie in [1,3]. Hence conclude that the conditions of the Fixed Point Theorem are not satisfied. (c) However, by plotting y = 9₁(x) and y = x on the same axis, observe that g₁ has a unique fixed point in [1,3].
Chapter9: Quadratic Equations And Functions
Section9.6: Graph Quadratic Functions Using Properties
Problem 9.104TI: A path of a toy rocket thrown upward from the ground at a rate of 208 ft/sec is modeled by the...
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Methods of Numerical Analysis:
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![Problem 1:
(a) Show that
Let f(x) = x³ + 5x² - 4x - 20
91(x) = 1x³
5
+ = 2² 5
4
92(x) = √√ (20 + 4x − x³)
-
both have fixed points p when f(p) = 0 (i.e. a fixed point that is a root of f).
(b) Find a value of g₁(x) that does not lie in [1,3]. Hence conclude that the conditions
of the Fixed Point Theorem are not satisfied.
(c) However, by plotting y = 9₁(x) and y = x on the same axis, observe that g₁ has a
unique fixed point in [1,3].
(d) Use the EVT to show that g₂ (x) € [1,3] for all x € [1, 3].
(e) Either by showing it algebraically, or from the graph (you don't have to submit the
graph), show that g2₂(x) = c for at least one x € [1, 3] for some explicit c≥ 1. You
need to find a valid value of x and an accompanying c.
(f) Even in light of (c), can we conclude that the Fixed Point Theorem will converge to
a root of f using 9₁?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f3fcee3-df4f-43f5-a70c-56c5ba155c30%2F606ca4e8-c04d-44a9-9ec2-b6cfb515db17%2Fk6b2brm_processed.png&w=3840&q=75)
Transcribed Image Text:Problem 1:
(a) Show that
Let f(x) = x³ + 5x² - 4x - 20
91(x) = 1x³
5
+ = 2² 5
4
92(x) = √√ (20 + 4x − x³)
-
both have fixed points p when f(p) = 0 (i.e. a fixed point that is a root of f).
(b) Find a value of g₁(x) that does not lie in [1,3]. Hence conclude that the conditions
of the Fixed Point Theorem are not satisfied.
(c) However, by plotting y = 9₁(x) and y = x on the same axis, observe that g₁ has a
unique fixed point in [1,3].
(d) Use the EVT to show that g₂ (x) € [1,3] for all x € [1, 3].
(e) Either by showing it algebraically, or from the graph (you don't have to submit the
graph), show that g2₂(x) = c for at least one x € [1, 3] for some explicit c≥ 1. You
need to find a valid value of x and an accompanying c.
(f) Even in light of (c), can we conclude that the Fixed Point Theorem will converge to
a root of f using 9₁?
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