Rationals
Find the Vertical and Horizontal Asymptotes of:
f(x)= (4x+1)/(x-2)
VA comes from denominator of (x-2). Cannot have zero in the denominator, so x can't equal 2.
HA comes from ratio of 4x/x, since degree of x is the same, look at coefficients of 4/1.
VA: x=2 and HA: y=4
True or False: A parabola, such as y=x2, is a one-to-one function.
False. A parabola of y=x2 is not a one-to-one function because it fails the horizontal line test.
In order to make it a one-to-one, the domain would need to be restricted, such as y=x2, for x>0.
If g(x)= 5/(x-1), and h(x)= 2x-3, find g(h(1)).
g(x)= 5/(x-1), h(x)= 2x-3
for g(h(1)), first find h(1)
h(1)= 2(1)-3
h(1)= -1
g(h(1))= 5/ [(-1)-1]
g(h(1))= 5/(-2)
g(h(1))= -5/2
What are the transformations on the function:
f(x)= 2x+1 -4
f(x)= 2x+1 -4
The graph has been shifted left 1 unit [from the (x+1)] and shifted down 4 units [from the -4].
log2 256 = ?
log2 256 = ?
change of base
(log 256)/(log 2) = 8
log2 256 = 8
--Or--
Rewrite: log2 256 = x ---> 2x = 256
2x = 28
x=8, so log2 256 = 8
Find all Asymptotes of:
f(x)= 2x/(x2-1)
f(x)= 2x/(x2-1)
For VA, what makes zero in denominator? 1 & -1
VA: x= 1, -1
For HA, ratio of 2x/x2, "bigger below...y=0"
HA: y=0
Find inverse of h(x)= -3x+2
h(x)= -3x+2
y = -3x+2
x = -3y+2
x-2 = -3y
y = (x-2)/-3
h-1(x) = -(x-2)/3
If h(x)= 2x-3, and f(x)= \/(3-x), find h(f(x)).
h(x)= 2x-3, f(x)= \/(3-x)
h(f(x)) = h(\/(3-x))
= 2(\/(3-x))-3
h(f(x)) = 2\/(3-x)-3
The parent function is f(x)= 4x. The function g(x) has been transformed from this parent function by being shifted right 2 units and shifted up 1 unit.
Write the function g(x).
f(x)= 4x, g(x) is right 2, up 1
g(x)= 4x-2 +1
Find the domain of g(x)= ln(3-2x)
g(x)= ln(3-2x)
set argument greater than zero & solve:
3-2x > 0
x < (3/2)
Domain: (-infinity, 3/2)
What is the Horizontal Asymptote and does the graph cross it at any point?
f(x)= (3x2+8x-5)/(x2+3)
f(x)= (3x2+8x-5)/(x2+3)
HA: degree same of (3x2)/(x2) so 3/1, y=3
Substitute 3 in for y:
3 = (3x2+8x-5)/(x2+3)
3x2+9 = 3x2+8x-5
14 = 8x
x = 7/4
The graph crosses the HA at the point (7/4, 3).
The p(x) has a domain of [0, infinity) and a range of [1, infinity).
What are the domain and range of p-1(x)?
Since p-1(x) is the inverse of p(x), the domain and range will be switched.
p-1(x) Domain is [1, infinity) and Range is [0, infinity).
The q(x)= -2x, and r(x)= \/(1-x)
Find (q*r)(x) and (q0r)(x).
The q(x)= -2x, and r(x)= \/(1-x)
(q*r)(x)= q(x) * r(x)
= (-2x)*(\/(1-x))
(q*r)(x)= -2x\/(1-x)
(q0r)(x)= q(r(x))
=q(\/(1-x))
= -2(\/(1-x))
q(r(x))= -2\/(1-x)
Jerome wants to invest $26,000 for his retirement. He can invest either at 5.1% simple interest for 40 years or at 3.8% continuous compounding interest for 40 years. Which option results in the most total interest?
$26,000 (P) for 40 years (t)
5.1% Simple interest: I=Prt
I= (26000)(0.051)(40)
I= $53,040 (interest only)
3.8% Continuous compounding interest: A=Pert
A=(26000)e(0.038*40)
A= $118877.86 (total amount)I= Amount - Principle --> I= 118877.86 - 26000
I= $92,877.86 (interest only)
The simple interest results in $53,040 while the continuous results in $92,877.86, so the continuous compounding interest at 3.8% is the better option.
Condense/Write as one log:
4log5y - 3log5x - 1/2log5z
4log5y - 3log5x - 1/2log5z
= log5y4 - log5x3 - log5z1/2
= log5y4 - log5x3 - log5(\/z)
= log5y4 / log5x3(\/z)
= log5[y4 / x3(\/z)]
Is there a hole in the function?
f(x)= (x2-x-2)/(x-2)
f(x)= (x2-x-2)/(x-2)
Factor numerator: f(x)= (x-2)(x+1)/ (x-2)
The (x-2) reduce/cancel each other, meaning the value that makes zero here (x=2) will cause problems.
The rest of the function is f(x)= x+1, so substitute the problem x-value in to find the problem y-value.
f(x)= x+1 --> y= (2)+1 --> y=3
These two values are a location. The hole in the graph will be a the point (2,3).
Find the inverse of n(x)= (x-4)/(x+2)
n(x)= (x-4)/(x+2)
y = (x-4)/(x+2)
x = (y-4)/(y+2)
x(y+2) = y-4
xy+2x = y-4
xy-y = -2x-4
y(x-1) = -2x-4
y= (-2x-4)/(x-1)
n-1(x) = (-2x-4)/(x-1)
f(x)= x3-3x, and g(x)=\/(2x). Find the f(g(x)) and the domain.
f(x)= x3-3x, g(x)=\/(2x)
f(g(x))= f(\/(2x))
= (\/(2x))3 -3(\/(2x))
= 2x\/(2x) -3\/(2x)
f(g(x))=2x\/(2x) -3\/(2x)
Domain Inside function, g(x):
2x >/= 0 ---> x>/=0 ---> [0, infinity)
Domain Final function, f(g(x)):
restrictions are the same as g(x)
Domain: [0, infinity)
Solve and leave as exact answer: 32x+1 = 43x
32x+1 = 43x
ln 32x+1 = ln 43x
(2x+1) ln3 = (3x) ln4
2xln3 + ln3 = 3xln4
ln3 = 3xln4 - 2xln3
ln3 = x (3ln4 - 2ln3)
x = ln3 / (3ln4 - 2ln3)
Solve:
ln(x) + ln(x+2) = ln(x+6)
ln(x) + ln(x+2) = ln(x+6)
ln(x(x+2)) = ln (x+6)
ln(x2+2x) = ln(x+6)
x2+2x = x+6
x2+x-6 =0
x= -3, 2
Check in original equation: -3 extraneous
x ={2}
Find the slant (oblique) asymptote of:
h(x)= (-3x2+4x-5)/(x+6)Can do either long division or synthetic, but here's synthetic:
-6| -3 +4 -5
____18__-132
-3 22 |-137
toss the remainder of -137
y= -3x+22
The slant asymptote is y= -3x+22
Check if h(x) and h-1(x) are inverses of each other.
h(x)= -3x+2, and h-1(x)= -(x-2)/3
h(x)= -3x+2, h-1(x)= -(x-2)/3
h(h-1(x))= h(-(x-2)/3)
= -3(-(x-2)/3)+2
= -1(-x+2) +2
= x-2+2
h(h-1(x))= x
h-1(h(x))= h-1(-3x+2)
= -[(-3x+2)-2] / 3
= (3x -2 +2) /3
h-1(h(x))= x
Yes they are inverses
The f(x)= 4/(x2-9), and g(x)= \/(3-x). Find the (f0g)(x) and the domain.
f(x)= 4/(x2-9), g(x)= \/(3-x)
(f0g)(x)= f(g(x))
=f(\/(3-x))
= (4)/[(\/(3-x)2 -9]
= (4)/ (3 -x -9)
= (4)/ (-x-6)
f(g(x)) = - 4/(x+6)
Domain Inside function g(x):
3-x >/=0 --> x</= 3 --> (-infinity, 3]
Domain final function f(g(x)):
x+6 =/=0, so x can't be -6
Combine 2 restrictions of less than/equal to 3 and not -6:
Domain f(g(x)): (-infinity, -6)U(-6, 3]
Strontium-90 has a half-life of approximately 28.9 years. If 15 micrograms is present in a sample, use the function to evaluate. Round to 3 decimals.
A(t)= 15(1/2) t /28.9
Find a) amount after 28.9 years and b) amount after 289 years.
A(t)= 15(1/2) t /28.9
a) t=28.9
A(28.9)= 15(1/2) 28.9 /28.9
A(t)= 7.5 micrograms (half of original 15 mcg)
b) t= 289
A(289)= 15(1/2) 289 /28.9
A(t)= (approx) 0.015 micrograms
Solve:
log2 (x) -2 = - log2(x+3)
log2 (x) -2 = - log2(x+3)
log2 (x) + log2(x+3) = 2
log2 (x(x+3)) = 2
log2 (x2+3x) = 2
x2+3x = 22
x2+3x = 4
x2+3x-4 =0
x= 1, -4
Check in original equation: -4 is extraneous
x = {1}