e. f: R→ [-1, 1], f(x) = cos x f. f: R→R, f(x) = |x|. g. 9: S¹ x I→ R², g(x, y) = x. (Here, it is understood that x E S¹ and ye I.) h. h: S¹→ R², h(x, y) = (x, y). Pro
e. f: R→ [-1, 1], f(x) = cos x f. f: R→R, f(x) = |x|. g. 9: S¹ x I→ R², g(x, y) = x. (Here, it is understood that x E S¹ and ye I.) h. h: S¹→ R², h(x, y) = (x, y). Pro
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
E,f,h,g
![e.
01 (s.))
0= x 1 (010)
= (x)}
S=T=R,
f(x) = x² - 1.
ЕЯ+Я:
Problem 7. For each function, give an explicit expression for its image, and
say whether the function is an injection, surjection, or bijection. If the function
is a bijection, also give its inverse.
a. f: RR, f(x) = x³
b. f: R→ R2, f(x) = (x,x³) [here, just use the bracket notation to express
im(f).]
c. f: R→R, f(x) = x²
d. f: R→ [0, ∞), f(x) = x²
e. f: R→ [-1, 1], f(x) = cos x
f. f: R→R, f(x) = |x|.
g. 9: S¹ x I→ R2, g(x, y) = x. (Here, it is understood that x E S¹ and
ye I.)
h. h: S¹→ R², h(x, y) = (x, y).
Problem o](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3bf7da54-5138-4b57-a0f9-aec30fa8e8cf%2F231b3d64-cae8-49c4-a10f-fa36fc274eaa%2Frz7xfn9r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:e.
01 (s.))
0= x 1 (010)
= (x)}
S=T=R,
f(x) = x² - 1.
ЕЯ+Я:
Problem 7. For each function, give an explicit expression for its image, and
say whether the function is an injection, surjection, or bijection. If the function
is a bijection, also give its inverse.
a. f: RR, f(x) = x³
b. f: R→ R2, f(x) = (x,x³) [here, just use the bracket notation to express
im(f).]
c. f: R→R, f(x) = x²
d. f: R→ [0, ∞), f(x) = x²
e. f: R→ [-1, 1], f(x) = cos x
f. f: R→R, f(x) = |x|.
g. 9: S¹ x I→ R2, g(x, y) = x. (Here, it is understood that x E S¹ and
ye I.)
h. h: S¹→ R², h(x, y) = (x, y).
Problem o
Expert Solution
Step 1
(e) f : R → [-1, 1], f(x) = cos x
For every x € [-1, 1], we have preimage y in R such that f(y) = cos(y) = x.
In fact there are many pre-images given a value x in [-1, 1], one being cos-1(x) within [0, π].
Hence, image(f) = [-1, 1].
f is surjective.
But, we see, 0 ≠ 2π, but f(0) = f(2π) = 1.
There are many more such examples.
Hence, f is not injective.
Step by step
Solved in 4 steps
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