Functions and their inverses
Is this a function:
(x-5)2 + (y+3)2 = 9No.
Are these lines perpendicular, parallel, or neither:
y1 = (2/5)x - 1
y2 = 5x + 2y = 27
Perpendicular
What is the initial value of:
20e0.3t
20
180o = __ radians
pi
Domain / Range of:
f(x) = cos(x)
Domain: (-inf, inf)
Range: [1, 1]
g(x) = 4x-3/2 - 18x5
g-1(g(x)) = ???
x
Find the vertical / horizontal asymptotes and removable discontinuities:
(x3 + x2 - 12x)/(x3 - 9x)
Discontinuities: at x = 0, x = 3
Vertical Asymptote: x = -3
Horizontal Asymptote: y = 1
log2 (x) - log2 (3) + log2 (y)
Condense it.
log2((xy)/3)
Coterminal angle of -362o
358o
sin (3pi/4)
sqrt(2)/2
Vertical stretch: 5/2
Horizontal translation: 3 units right
Vertical translation: 48 units up
f(x) = (1 - 2x5)3
End Behavior: x -> inf, y -> -inf
x -> -inf, y -> inf
Leading Term: -8x15
What is the next step to the problem:
log3(3x + 6) = 81
Apply 3^x to both sides to get:
3x + 6 = log3(81)
If my given solutions are pi/6 and 2pi/6 in 0 < x < 2pi, what do I add to my solutions to find solutions in a new domain, such as 4pi < x < 6pi
You would add 2pi(k) to both solutions where k is any integer (0, 1, 2, 3)
Solve:
cot^-1(cot(sin(-11pi/6)))
= sin(-11pi/6)
2pi - 11pi/6 = pi/6
sin(pi/6) = 1/2
Domain of:
(x2 - 2x + 6)3
----------------
(x-6)(x2+9)
(-inf, 6) U (6, inf)
How does the multiplicity of the zeros affect the graph of y(x)?
y(x) = (x - 6)2(x + 5)
Bounces at (6, 0). Goes through (-5, 0).
Find the x-int, y-int:
f(x) = 3(4x) - 19
x-int: (log4(19/3), 0)
y-int: (0, -16)
Radians of first quadrant [0o,90o]
0, pi/6, pi/4, pi/3, pi/2
Find
cos^-1(-sqrt(2)/2)
x = 3pi/4 and 5pi/4
f(x) = 2x2 + 7x2/3
g(x) = 3x3 + 7
f(g(x)) = ???
2(3x3+7)2 + 7(3x3+7)2/3
2x2 - 9x + 7 = 0
Find x.
x = (9/4) +- (5/4)
log7 (49x) = log3 (1/243)
x = -5/2
Solve: cot (sin (pi(cos2(pi/2) + sin2(pi/2))))
undefined
cot(x) / csc(x)
cos(x)