2. The domain of the function f(x) = the square root of 4-x^2 is
A. x < -2 or x>2
B. x/= 2
C. -2
We learned how to do this in advanced math, but I forgot. To find the domain, set what is inside the square root equal to zero then solve for x. Once you have x, set up intervals and then check the intervals to see if you get a real number.
4-x^2=0
x=2, -2
(-infinity, -2) (-2,2) (2, infinity)
the square root of 4-(-3)^2 = imaginary number
the square root of 4-(0)^2 = + or - 2
the square root of 4-(3)^2 = imaginary
Therefore, your domain is -2
7. Find k so that f(x) x^2 - 16 / x-4 x does not =4
k x=4 is continuous for all x.
A. All real values of k make f(x) continuous for all x.
B. 0
C. 16
D. 8
E. There is no real value of k that makes f(x) continuous for all x.
For this problem, take the limit as x->4.
(x+4) (x-4) / (x-4)
The (x-4)s cancel and you are left with x+4
Plug in 4 and you get 4+4 = 8
The answer is D. 8.
12. Find a positive value c, for x, that satisfies the conclusion of the Mean Value Theorem for Derivatives for f(x)=3x^2-5x+1 on the interval [2,5].
A. 1
B. 13/6
C. 11/6
D. 23/6
E. 7/2
The Mean Value states f(b)-f(a) / b-a, so all you have to do is plug into that formula then set that equal to the derivative.
3(5)^2-5(5)+1 - [3(2)^2-5(2) +1] / 5-2
75-25+1 - [12-10+1] / 3
51-3/3
48/3 = 6x - 5
48= 18x - 15
63 = 18x
x = 7/2 E.
45. If f(x) is continuous and differentiable and f(x) = ax^4+5x ; x bx^2-3x; x>2 then b=
A. 0.5
B. 0
C. 2
D. 6
E. There is no value of b.
The first step is to set the equations equal and plug in 2 for x.
a(2)^4 + 5(2) = x(2)^2-3(2)
16a+10 = 4b -6
Then take the derivative of each, plug in 2, and set them equal.
4ax^3 +5 = 2bx -3
4a(2)^3 + 5 = 2b(2) - 3
32a+5 = 4b-3
Now solve each equation for a.
16a+10=4b-6
a= 4b-16/16
a=b-4/4
32a+5=4b-3
a= 4b-8/32
a= b-2/8
Next, set the two equations solved for a equal and solve for b.
b-4/4 = b-2/8
8b-32= 4b - 8
4b=24
b=6 D.
I'm having the most trouble with substitution, and I can use a review on optimization.
Optimization
ReplyDeleteOptimization can be used for anything from finding the maximum amount of fencing to make a pen to finding the least amount of volume for a cylindrical cone. This concept is used commonly throughout the world and needs to be mastered for college level mathematics.
Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers.