False. Since (2,1) and (2,2) have the same
x-coordinate and different y-coordinates, the vertical
line test fails.
True. square_root((4-1)^2+(-2-2)^2) = 5.
True. The domain and range of f^{-1}(x) are
the range and domain of f(x).
True. By the Rational Root Theorem the
potential rational zeros of
x^6-10x^2+1 are +/-1, but neither is a zero.
True. This is a consequence of the Fundamental
Theorem of Algebra.
False. If x=1 and y=4, then the left side equals 1
while the right side equals 0.
Complete the sentence.
The function f(x) is an even function
if it satisfies f(x)=f(-x). Equivalently, the graph of
y = f(x) is symmetric about the y-axis.
The graph of y=f(x+1) is obtained from the
graph of y=f(x) by shifting left one unit of distance.
A function has an inverse if and only if
it is one-to-one.
The graph of y=f^{-1}(x) can be obtained
from the graph of y=f(x) by reflection through y=x.
The line is (y+3) = (-1/2)(x-2).
Putting x^2+4x+y^2-8y+11=0 into standard form,
(x+2)^2 + (y-4)^2 = 3^2,
we see that the center of this circle is at (-2,4)
and the radius is 3.
The vertex of y = 3x^2 + 2x + 1 =
3(x+1/3)^2 + 2/3 is (x,y) = (-1/3,2/3).
We want to maximize xy subject to x + 2y = 20.
Thus we need to find the maximum of P = (20-2y)y = -2y^2 + 20y
= -2(y-5)^2 + 50. From this form it is easy to see that
the maximum of P is 50, and it occurs when y = 5 (and so x = 10).
Your sketch should indicate that:
there are vertical
asymptotes at x = -1, x = 1,
there is no sign change near either vertical asymptote,
as x --> +infinity, then y --> +infinity,
and as x --> -infinity, then y --> -infinity,
there is an oblique asymptote with equation y = 2x,
the only x-intercept is at x = 0,
the curve is tangent to the
axis at x = 0 and changes sign there,
the curve increases between x = -1 and x = +1.
Write down the inverses of the following functions:
[f(x) = 2x-7] f^{-1}(x) = (1/2)(x+7)
[f(x) = x^3 + 1] f^{-1}(x) = cube_root(x-1)
[f(x) = 2^x] f^{-1}(x) = log_2(x)
Possible answers are:
Since B = Pe^(rt) = Pe^(.1t) and we want B = 2P,
we have 2 = e^(.1t). Therefore t = 10ln(2) = 6.93147... years. OR
By the rule of 70,
the length of time until the money doubles is 70/10 = 7 years.