I KNOW IT IS LONG BUT ITS SOO IMPORTANT PLEASE HELP!

Question

So we have this table to as an example to help us throughout the questions

Student Pair	Function one	Function 2	Composition
1	g(x)=1/x	f(x)=(2x^2)-8	g(f(x))
2	k(x)=sqrt(x)	h(x)= x-1	k(h(x))
3	L(x)= |x|	m(x)= -x^2	L(m(x))
4	t(x)=2 ^ (x+1)	s(x)=log_2 (x)	t(s(x))

1. Choose two of the pairs of functions and the associated composition. Complete the table for the given values of x.

x	Function 2	Composition Function1(Function2(x))
-3
-2
-1
0
1
2

2.Make a conjecture (your best guess) for the domain of your composition. Explain how you made this conjecture.

3.Make a conjecture for the range of your composition. Explain how you made this conjecture.

4.Write the algebraic representation or formula for this composition. Use this function formula to complete the table

x	Composition Function1(Function2(x))
-3
-2
-1
0
1
2
3

How do the values in this table compare to the table you completed in part 1?

5.Which table do think is easier to complete, the table in part 1 or in part 4? Explain why.

6.Use your graphing calculator to graph the composition you selected in part 1:

7.

a. Sketch the graph of this composition. How does this graph support or contradict the domain and range you identified in parts b and c?

b. Consider the functions w(x)=ln⁡(x−1) and y(x)=ex+1

c. Create the composition w(y(x)).

d. Graph the two functions w(x) and y(x)on the same screen. What do you notice about these two graphs?

Answer 1

Ok, so in this question you are asked to demonstrate your ability to evaluate composite functions, and to determine domain and range for various function types. As an example I will do the second composite function, k(h(x)), where h(x) = x-1 and k(x) = sqrt(x).

Remember that a function is an input --> output "machine"; we put a value in and another value comes out. A composite function is like two machines in a factory. In our example the output of h(x) becomes the input for k(x).

So if we put x = 5 in to h(x); called function 2 in this assignment; we will get h(5) = 5-1 = 4. Now we take that output and use it as input for g(x); ie function1; giving g(4) = sqrt(4) = 2. So overall k(h(5)) = k(4) = 2. This is easiest to do when the input numbers are one higher than a square number, but we can evaluate the composite function for other x values using a calculator.

The assignment asks us to evaluate h(x) for integer values of x from -3 to +2, so we should get integer values from -4 to +1 (one less than our input values). We would then put those values into the table, in the column labeled function 2, because they are the values for function 2, ie h(x).

Next we put those new values into k(x), and we start to understand why they are asking about dominion and range.

Linear functions like h(x) = x-1 have a domain and range that include all real numbers, but the square root function cannot be applied to negative numbers, so its domain - the set of possible x values - starts at 0, and its range - the set of all possible y values - also includes only zero and all positive real numbers. The square root function is defined to return the positive square root That is we can put in any non-negative real number and get out any non negative real number.

But they are asking about the domain and range of the composite function, so we need to think about the input to the first function and the output from the second. We see from our chart that putting +1 into h(x) gives h(1) = 0, while putting in zero gives h(0) = -1, so the domain of the composite function is x >=1, since putting any number less than that into h(x) will give a negative number as the input to k(x). The range remains y >=0, since the input to k(x) includes its entire domain.

To complete the assignment we need to fill in values for the composite function. I would write undefined for each x that is outside the domain (x = -3 to 0), then give the values from a calculator for the square roots, or just write them with a radical sign if needed, eg k(h(3)) = k(3-1) = sqrt(2).

The algebraic form of the composite is just putting the algebraic form of the second function in place of x f=or the first function, and simplifying as needed: k(h(x) = sqrt(x-1), which cannot be simplified. Evaluating for each x should give the same result as in the first table.

I hope that helps you understand what composite functions are, and how to evaluate them.

Remember to think about the domain and range for the "second function" or inside function first, since the x input actually goes into that "machine" first, and the output (range) of the inner function becomes the input (so may restrict the domain of) the outer function, which could potentially limit its output range.

Composite functions are not that scary if you think of them as a series of input --> output machines. You can combine a big string of them and make a whole factory!