Find f ( 1 ) , f ( 2 ) , f ( 3 ) , f ( 4 ) , and f ( 5 ) if f ( n ) is defined recursively by f ( o ) = 3 and for n = 0 , 1 , 2 , ... a) f ( n + 1 ) = − 2 f ( n ) . b) f ( n + 1 ) = 3 f ( n ) + 7 . c) f ( n + 1 ) = f ( n ) 2 − 2 f ( n ) − 2. d) f ( n + 1 ) = 3 f ( n / 3 ) .
Find f ( 1 ) , f ( 2 ) , f ( 3 ) , f ( 4 ) , and f ( 5 ) if f ( n ) is defined recursively by f ( o ) = 3 and for n = 0 , 1 , 2 , ... a) f ( n + 1 ) = − 2 f ( n ) . b) f ( n + 1 ) = 3 f ( n ) + 7 . c) f ( n + 1 ) = f ( n ) 2 − 2 f ( n ) − 2. d) f ( n + 1 ) = 3 f ( n / 3 ) .
Solution Summary: The author explains how to calculate f(n) if it is defined recursively.
A function is defined on the interval (-π/2,π/2) by this multipart rule:
if -π/2 < x < 0
f(x) =
a
if x=0
31-tan x
+31-cot x
if 0 < x < π/2
Here, a and b are constants. Find a and b so that the function f(x) is continuous at x=0.
a=
b= 3
Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
f(x) = (x + 4x4) 5,
a = -1
lim f(x)
X--1
=
lim
x+4x
X--1
lim
X-1
4
x+4x
5
))"
5
))
by the power law
by the sum law
lim (x) + lim
X--1
4
4x
X-1
-(0,00+(
Find f(-1).
f(-1)=243
lim (x) +
-1 +4
35
4 ([
)
lim (x4)
5
x-1
Thus, by the definition of continuity, f is continuous at a = -1.
by the multiple constant law
by the direct substitution property
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