11.) Inverses
F(x)
= 31,573,161x^2 + 7,621
Step
1: Set F(x)= to y= and then switch the x and y in the equation.
X
= 31,573,161y^2 + 7,621
Step 2: Solve for Y
X
= 31,573,161y^2 + 7,621
-7,621
-7,621
X – 7,621 = 31,573,161y^2
√(X - 7,621) = √(31,573,161y^2)
± √(X – 7,621) = 5,619y
± √(X – 7,621)/5,621 = (5,619y)/5,621
Y = ±√(X – 7,621)/5,621
Step 3: Use inverse notation.
F-1(X)= ±√(X –
7,621)/5,621
2.) Domain and range
√(609x+36) / √(906x+63)
Step 1: Set the numerator greater
than or equal to 0, and the denominator greater than 0, since you cannot divide
by 0.
609x+36≥0 906x+63≥0
Step 2: Solve the inequalities.
609x+36≥ 906x+63≥0
-36 -36 -63 -63
609x≥-36 906x≥-63
/609 /609 /906 /906
x≥-36/609 x≥-63/906
This reduces to D: [-12/203, ∞) This reduces to D: (-21/302, ∞)
DF: [-12/203, ∞)
R: [0, ∞)
33.)
Completing the square
Step 1: Subtract the “c” from both
sides.
5 = 487x^2 + 2000x + 79
-79 -79
-74 = 487x^2 + 2000x
Step 2: Factor out the “a” value
Step 3: Use the new “b” value and use
(b/2)^2 to find the perfect square “c” value.
-74 = 487(x^2 + 2000/487x+__)
+1,000,000/237,169
Step 4: Multiply the new “c” value by
the factored out “a” value and add it to the other side of the problem
-74 = 487(x^2 +
2000/487x+1,000,000/237,169)
+1,000,000/487
963,962/487 = (x^2 + 2000/487x +
1,000,000/237,169)
Step 5: Solve for X.
963,962/487 = 487(x + 1000/487)^2
/487 /487
963,962/237,196 = (x + 1000/487)^2
√(963,962/237,196) = √(x +
1000/487)^2
±√(963,962/237,196) = x + 1000/487
-1000/487 -1000/487
X = -1000/487 ± √(963,962/237,196)
This can be reduced to X = -1000/487
± √(963,962)/487
44.)
Farmer Splavin has 2,469,636 ft^2 of
fencing. He only needs three sides of fencing because there is a river of lava
that flows through his land. If Farmer Splavin does not fence alongside the
lava, what is he maximum area that the fencing can enclose.
Step 1: Use both perimeter and area
equations. Solve for Y in the perimeter equation, and plug the term into the
area equation.
P = 2x + y A = xy
2,469,636 = 2x + y A
= x (2,469,636 – 2x)
-2x -2x A = -2x^2 + 2,469,636
2,469,636 - 2x = Y (-b/2a)
= -2,469,636/2(-2)
X
= 617,409
Step 2: Use the new X value and plug
it into the perimeter equation that equals Y.
-2(617409) + 2,469,636 = Y
Y = 1234818
Step 3: Use the X value and the new Y
value and plug them into the area equation.
xy
= (7.623877466X10^11) sq ft.
55.)
Factoring polynomials
F(x) = 262,144x^3 - 1,953,125
Step 1: Use the cube root of the A and B values to get the values
needed for the factored form.
A = 3√(262,144x^3) B
= 3√(1,953,125)
A = 64x B
= 125
Step 2: Use the new values and plug them into the factor formula.
(A-B)(A^2 + AB + B^2) = (64x – 125)(4,096x^2 + 8,000x + 15,625)
66.)
Combining functions
Find F(G(H(7)))
F(x) = 469,752x^2 – 4
G(x) = 72x
H(x) = 672x^2 + 970x
Step 1: Plug 7 into the H(x) equation.
H(7) = 672(7)^2 + 970 = 39,718
Step 2: Plug 39,718 into the G(x) equation.
G(39,718) = 72( 39,718) = 2,859,696
Step 3: Use the new value and plug it into the F(x) equation for the
final answer.
F(2,859,696) = 469,752(2,859,696)^2 – 4 = (3.84156666X10^18)
Tyler's reflection: We chose the concepts we did to create this problem set because we believed they accurately depicted the problems we had to complete in each of the units. The problems range through all three units we've worked with and the most important problems we have completed. These problems provide and overview of my best mathematical understanding of what I have learned so far by how complex the problems were that we created and by our explanations of the problems to the point that an interested learner could understand and learn from us. Although I did not learn anything from this assignment, it did give me a chance to review all of the types of problems we have encountered and has helped me start preparing for the exam and future tests.
Edmund's reflection:
Our group ended up doing a few problems spanning over all the units that we have done this trimester. We picked the problems that we felt showed one's overall understanding of the entire unit. We felt that in the functions basics unit finding the inverse of functions, the domain and range, and combination of functions were the most important concepts to learn. In the quadratics unit, we picked a completing the square problem and a substitution problem(farmer Ted problem). In polynomials, since we are working it on now and are lacking in some knowledge of the unit we picked a factoring polynomials problem. Each of these problems were the most important in each unit. I learned quite a bit from this project, but mostly I was able to review most of what we have learned up to this point. We were also able to create difficult problems from the most important concepts, and it was excellent practice for all of us. The ability to teach all of the problems aided our understanding as well.
Ryan's reflection:
We chose problems that we have worked on throughout the trimester so far. The problems we chose represent each of the three units: Function Basics, Quadratics, and Polynomials. We did three problems from the Function Basics unit: finding the inverse, domain and range, and the combination of functions using the composition method. The step by step processes in these problems represent our understanding of the content so far. In the Quadratics unit, we did a farmer Ted problem, and a completing the square problem. We made these problems difficult, and it helped us understand how to solve and create a step by step process in order to show our full knowledge of the unit. Since we are still working on the polynomials unit, we chose to only do one example problem. These problems represented the units they were from well. I learned a lot from this project, simply because we chose to do more challenging problems. It helped me understand the concepts we've learned even better than before, and also to learn from any mistakes I was making.
No comments:
Post a Comment
Note: Only a member of this blog may post a comment.