Decreasing in: (-∞, 3] Increasing in: [3, +∞)
f(-x) = 2(-x)3 - (-x) = -2x3 + x = -(2x3 - x) = -f(x)Therefore, f(x) is an odd function.
For each value of x, there is only one possible value of y; therefore, thr right hand side of the equation defines a function.
For each value of x, there is only one possible value of y; therefore, right hand side defines a function.
Solving for y, we get:
______
y = ±√2x + 1
For each value of x, there is may be more than one value of y; therefore,
equation right hand side is not a function of x.
Solving for y, we get:
______
y = ±√4 - x2
For each value of x, there is may be more than one value of y; therefore,
equation is not a function of x.
For each value of x, there is only one value of y; therefore, list defines a function.
____
f(x) = floor( √ x3 )
x f(x)
----------
0 0
1 1
2 2
3 5
4 8
5 11
6 14
7 18
8 22
9 27
_______
g(x) = ceil( √ x + 4 )
x f(x)
----------
0 2
1 3
2 3
3 3
4 3
5 3
6 4
7 4
8 4
9 4
Domain is the set of real numbers: (-∞, +∞)
Range is the set: [-4, +∞)
All real numbers; x ≠ 2
All real numbers; x ≠ 2, x ≠ -2
Domain is the set of real numbers: (-∞, +∞) Range is the set: (0, +∞)
2x + k if x ≤ 1
f(x) =
x2 + 1 if x > 1
Need to solve 2x + k = x2 + 1 at the boundary, x = 1:
2 + k = 2
k = 0
$0.37 if 0 < x ≤ 1oz
$0.60 if 1 < x ≤ 2oz
$0.83 if 2 < x ≤ 3oz
$1.06 if 3 < x ≤ 4oz
p(x) = $1.29 if 4 < x ≤ 5oz
$1.52 if 5 < x ≤ 6oz
$1.75 if 6 < x ≤ 7oz
$1.98 if 7 < x ≤ 8oz
$2.21 if 8 < x ≤ 9oz
f(x) = x2 + 3x + 2
= (c + 1)2 + 3(c + 1) + 2
= c2 + 2c + 1 + 3c + 3 + 2
= c2 + 5c + 6
(x + h)2 - 2(x + h) + 1 - x2 + 2x - 1
g(x) = -------------------------------------
h
x2 + 2xh + h2 - 2x - 2h + 1 - x2 + 2x - 1
g(x) = -----------------------------------------
h
2xh + h2 - 2h
g(x) = --------------- = 2x - 2 + h
h
f(2)g(2) = 22×32 = 4×9 = 36 f(g(2)) = 29 = 512 g(f(2)) = 34 = 81
f(c)g(c) = c-1·(c3 - 1) = (c3 - 1)/c f(g(c)) = (c3 - 1)-1 = 1/(c3 - 1) g(f(c)) = (c-1)3 - 1 = (1/c3) - 1
_
f(x)g(x) = x2·√x
_
f(g(x)) = ( √x )2 = x
___
g(f(x)) = √ x2 = x
f(0) = 02 - 0 - 2 = -2 f(4) = 42 - 4 - 2 = 10Since the value of f(x) has opposite signs at the endpoints of the interval; the graph of f(x) must cross the x-axis.