Analee Abreu-Rodriguez 11/23/14
Algebra II – Mr. Tuetken
1.3 Module one quiz
Both Splot and Fizzle got the wrong answer. When Splot added 4 to both sides he was really supposed to subtract. That would mean 14 would go down to 10 and the equation would then be 5x = 10, which he then would have divided by 5 on both sides and gotten x = 2. Fizzle was doing okay until he subtracted 14 from 4, but only really took away 2 from 4 and got -12 instead of -10. Then instead of dividing -12 and -5x by -5, which still would have been wrong, he added 5 to both sides and got -7 = x instead of x = 2, which is what he would have gotten if he did the problem correctly. Which he would have done if he had subtracted correctly from 4, and then divided -10 by -5.
Suppose we have a function f(x)= x^2 + 6 * x
We would know this is a function because when graphed it passes the vertical line test. Meaning that when a vertical line is drawn on the graph it never crossed to points at the same time.
We need to use that function to find f(3). In order to do that we first have to substitute 3 for x in the equation f(x)= x + 6. The equation would go from f(x)= x^2 + 6 * x to f(3)= (3)^2 + 6 * 3, then you would multiply the 3 in the parenthesis by 3 and get 9, after that you multiply 6 by 3 to get 18 and add that to 9, which would mean f(3) would be 27.
In order to find the inverse of the function f(x)= x^2 + 6 * x, you first have to make f(x) into y, that way the function is y= x^2 +6 * x, then you have to switch y with x to get x = y^2 + 6 * y then subtract 6 from both sides to get x – 6 = y^2 * y,then you would divide by y and get x-6 /y = y^2 then you would use the square root, the function would be plus or minus the square root of x-6 over y, then the function will equal f^-1(x) which is the inverse. Which would mean the inverse is +- the square root of x – 6 over y = f^-1(x).
Okay, my new function h(x) will be h(x) = 9x and I will let f(x) stay the same as before, my number for x will be 4 so it will be h(4). Now, functions can be done two different ways, which is f(h(x)) and h(f(x)), with