Asked by Samantha
Okay, so I think these are right, but I would appreciate if someone could check them and tell me if something is wrong and what the right answer is. I'd also appreciate an explanation if possible. :)Thank you!
7. Given that
f(x)={x^3 if x ≥ 0
{x if x < 0
which of the following functions is even?
I. f(x)
II. f(|x|)
III. |f(x)|
I only
***II only
I and II only
I and III only
None of these
8. If f(x) is an odd function, which of the following must also be odd?
***–f(x)
f(|x|)
|f(x)|
f(x – 1)
None of these
9. If f(x) is an odd function, which of the following must be even?
-f(x)
f(|x - 1|)
| f(x)|
***f(x + 1)
None of these
10. If f(x) is an even function and g(x) is an odd function, which of the following must be even?
I. f(g(x))
II. f(x) + g(x)
III. f(x)g(x)
I only
II only
I and II only
II and III only
***I, II, and III
11. If f and g are odd functions, which of the following must also be odd?
I. f(g(x))
II. f(x) + g(x)
III. f(x)g(x)
I only
II only
I and II only
II and III only
I, II, and III
7. Given that
f(x)={x^3 if x ≥ 0
{x if x < 0
which of the following functions is even?
I. f(x)
II. f(|x|)
III. |f(x)|
I only
***II only
I and II only
I and III only
None of these
8. If f(x) is an odd function, which of the following must also be odd?
***–f(x)
f(|x|)
|f(x)|
f(x – 1)
None of these
9. If f(x) is an odd function, which of the following must be even?
-f(x)
f(|x - 1|)
| f(x)|
***f(x + 1)
None of these
10. If f(x) is an even function and g(x) is an odd function, which of the following must be even?
I. f(g(x))
II. f(x) + g(x)
III. f(x)g(x)
I only
II only
I and II only
II and III only
***I, II, and III
11. If f and g are odd functions, which of the following must also be odd?
I. f(g(x))
II. f(x) + g(x)
III. f(x)g(x)
I only
II only
I and II only
II and III only
I, II, and III
Answers
Answered by
Samantha
Oh, for 11 I chose A (first option)
Answered by
Steve
#7. None of these.
Plug in -x for x and see that the graphs are not symmetric about the origin. I mean really -- part of the graph is a straight line, and part is a cubic curve.
#8 ok
#9 f(x) is odd, so f(-x) = -f(x)
|f(-x)| = |-f(x)| = |f(x)|
so, even
f(x+1) is the same graph shifted left by 1, so it is no longer symmetric about the origin.
#10
Please use a little common sense.
Pick f(x) = x^2
g(x) = x
f+g = x^2+x, neither even nor odd
fg = x^3, odd
so, we are left with
f(g(-x)) = f(-g(x)) = f(g(x)), even
#11
f(-x)+g(-x) = -f(x)-g(x) = -(f+g), so odd
f(g(-x)) = f(-g(x)) = -f(g(x)), so odd
f(-x)g(-x) = -f(x) * -g(x) = f*g, so even
Answer is I and II only
Plug in -x for x and see that the graphs are not symmetric about the origin. I mean really -- part of the graph is a straight line, and part is a cubic curve.
#8 ok
#9 f(x) is odd, so f(-x) = -f(x)
|f(-x)| = |-f(x)| = |f(x)|
so, even
f(x+1) is the same graph shifted left by 1, so it is no longer symmetric about the origin.
#10
Please use a little common sense.
Pick f(x) = x^2
g(x) = x
f+g = x^2+x, neither even nor odd
fg = x^3, odd
so, we are left with
f(g(-x)) = f(-g(x)) = f(g(x)), even
#11
f(-x)+g(-x) = -f(x)-g(x) = -(f+g), so odd
f(g(-x)) = f(-g(x)) = -f(g(x)), so odd
f(-x)g(-x) = -f(x) * -g(x) = f*g, so even
Answer is I and II only
Answered by
Steve
Bad logic above:
#7. Graph is part line and part curve, so no axis of symmetry
#9. f is odd, so f(x+1) is just an odd function shifted left. Still no axis of symmetry.
#7. Graph is part line and part curve, so no axis of symmetry
#9. f is odd, so f(x+1) is just an odd function shifted left. Still no axis of symmetry.
Answered by
Antidisestablishmentarianism
no answer after 7 years wow.
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