The domain of f(x)=(1)/(sqrt(x^2-6x-7)) is
(1, 7)
[-1, 7]
x > -1 or x < 7
***{x < -1}U{x > 7}
(-∞, -1]U[7, ∞)
2. In which of the following is y a function of x?
I. y^2=9-x^2
II. |y|=x
III. y=(sqrt(x^2))^3
I only
II only
III only
***I and III only
I, II and III
3. You would probably use calculus to determine the area for which of the following shapes?
***(infinity symbol)
(rectangle)
(triangle)
(arrow)
(plus sign)
4. If g(x)=3x−1 and f(x)=(sqrt(9-x^2)) , then which of the following will have a domain of: [−3, 1/3) ∪ (1/3, 3]?
(g + f)(x)
(f/g)(x)
(g − f)(x)
***g(f(x))
f(g(x))
5. Which of the following is always true of odd functions?
I. f(-x)=-f(x)
II.f(|x|) is even
III. |f(x)| is even
All of these are true.
None of these are true.
I only.
II only.
***I and III only.
6. If f(x)= 3/(x-3), then f^-1(x)=
(x-3)/3
(x+3)/3
***3/(x+3)
3/(x-3)
No inverse exists.
7. The function f(x) = −|x − b| intersects g(x) = 2(x − 4)^2 at exactly one point. The value of b is:
-5
***-4
5
4
Any real number
8. If a right circular cone must have a volume of 12 cubic meters, then a function that gives the possible radii as a function of the height is:
r=12/pi*h
r=+-(sqrt(36/pi*h))
***r=(sqrt(36/pi*h))
h=(sqrt(36/pi*r^2))
r=36/(pi*h)
#1 ok
#2 III only
for y to be a function of x, each x must map to only one y.
#3 ok
#4 (f/g)(x)
f has domain [-3,3]
g has domain x≠1/3
#5 all of the above
f(|-x|) = f(|x|)
#6 Hmmm. I get 3(x+1)/x = 3/x + 3
That's your choice without the parentheses
#7 +4
#8 ok
his answers are right
Thank you!! Unfortunately I couldn't wait for your response, but I went over my answers before checking to see if they were right, and I got them all right! Thank you so much for your help! It means the world to me, and will count on you if I need help on more topics in Calculus. :)
Hey Samantha, were his answers right or were yours?
1. The domain of f(x) = (1)/sqrt(x^2-6x-7) is {x < -1}U{x > 7} because the function is undefined when the denominator becomes zero, which happens when x^2-6x-7 = 0. Solving for x, we find the roots x = -1 and x = 7, so we exclude these values from the domain.
2. The values of y in which y is a function of x are I and III only. In I, y^2 = 9 - x^2, we can solve for y as y = +/- sqrt(9-x^2), which means that for any given x, there are two possible values of y. In II, |y| = x, we can also have multiple values of y for a single x, such as y = x and y = -x. In III, y = (sqrt(x^2))^3 simplifies to y = x^3, which is a function.
3. Calculus would be used to determine the area of the (infinity symbol) shape. This shape has a loop and curves, which require integration techniques to find the area.
4. The function g(f(x)) will have a domain of [−3, 1/3) ∪ (1/3, 3]. In g(f(x)), we substitute f(x) = sqrt(9-x^2). Since the domain of f(x) is [-3, 3], we need to find the domain of g(f(x)) that is within this range. Solving for x in sqrt(9-x^2) = 3x-1 gives us x = -1/3 and x = 1/3 as the points where the domain changes. So, the final domain is [−3, 1/3) ∪ (1/3, 3].
5. The statement "f(-x) = -f(x)" is true for odd functions, but the statement "f(|x|) is even" is not always true for odd functions. However, the statement "|f(x)| is even" is true for odd functions. So, the correct option is I and III only.
6. If f(x) = 3/(x-3), then f^-1(x) = 3/(x+3) because to find the inverse of a function, we interchange x and y and then solve for y.
7. The value of b that makes f(x) = −|x − b| intersect g(x) = 2(x − 4)^2 at exactly one point is -4. The absolute value function, |-4 − b|, intersects the quadratic function 2(x − 4)^2 at one point when b = -4.
8. The function that gives the possible radii as a function of the height for a right circular cone with a volume of 12 cubic meters is r = (sqrt(36/pi*h)). The volume of a cone is given by V = 1/3 * pi * r^2 * h, and since we know the volume (12 cubic meters), we can solve for r in terms of h by rearranging the equation. The correct option is r = (sqrt(36/pi*h)).