5. (18 pts) Circle True if the statement is ALWAYS true;
circle False otherwise. No explanations are required.
Let f be a one-to-one function and f
its inverse.
(a)
If the point (2, 5) lies on the graph off,
then the point (5,2) lies on the graph of f.
True
False
f (2) = 5
f-1 (5) = 2
(b)
sin-1 (T) = 0
True
False
Since >1, ris not in the domain of sin-?; |
or
sin-1 (7) #0, since sin (0) = 0 #
Iff and g are two functions defined on (-1, 1), and if
(c)
limg (x) = 0, then it must be true that
lim[ f(x) g (x)] = 0.
True
False
Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x.
Then lim g (x) = lim (x) = 0, but
Lim ( f (x) g(x)] = lim [() x] = 18 0.
Iffis continuous on (-1, 1),
and if f (0) = 10 and lim g (x) = 2,
(a)
True
False
then lin (= 5.
Because lim : (*) - Limx-of (*) .f(0) = 20 = 5
If f is continuous on [1, 3], and if
f(1) = 0 and f(3) = 4, then the equation f(x) =
(e)
has a solution in (1, 3).
True
False
Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3),
by the Intermidiate Value Theorem there exists a
cin (1, 3) such that f (c) = .
(f)
Let f be a positive function with vertical
asymptote x = 5. Then
lim f (x) = +0.
True
False
Let f (x) = 1, if x < 5, and f (x) ==, ifx>5.
Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote.
5
-
X
So, f is a positive function with vertical asymptote x = 5 but
lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1,
which means that lim f (x) does not exist.
