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Section 2.3 #23: Find (fg)(x) and (gf)(x) and the domain of each:
f(x)=415x;g(x)=1x. Solution: First compute (fg)(x)=f(g(x))=f(1x)=415x=4xx5.

The domain of (fg) must lie in the domain of g, because we are first coputing g(x) and then taking f of g(x). The domain of g is all real numbers except x=0. Since the domain of f is all real numbers except 15 (set the denominator of f equal to zero to see this), we must also throw out any value of g(x) such that g(x)=15. This equation is 1x=15, which means x=5. Therefore the domain of (fg) is all real numbers except 0 and 5.

Now compute (gf)(x)=g(f(x))=g(415x)=1415x=15x4. The domain of f is all real numbers except for x=15 and the domain of g is all real numbers except x=0. Consequently, we must consider for which values of x is f(x)=5, i.e. 415x=5. Multiply by 15x to get 4=525x implying 25x=1, or x=125. Therefore the domain of (gf) is all real numbers except for 15 and 125.
Section 2.3 #40: Find f(x) and g(x) such that h(x)=(fg)(x), where h(x)=3x28. In other words, "decompose" the function h.
Solution: One way to proceed is to notice that the "outer" function is the cube root function and the "inner" function is x28. So we can take f(x)=3x and g(x)=x28 so that (fg)(x)=f(g(x))=f(x28)=3x28.

Section 2.4 #2: Determine visually whether the graph is symmetric with respect to the x-axis, the y-axis, or the origin:

Solution: It is symmateric with respect to the y-axis.

Section 2.4 #27: Find the point that is symmetric to the point (5,6) with respect to the x-axis, the y-axis, and the origin.
Solution: The point symmetric to (5,6) about the x-axis is the point (5,6). The point symmetric to (5,6) about the y-axis is (5,6). The point symmetric to the origin is (5,6).

Section 2.4 #40: Determine whether the function f(x)=7x3+4x1 is even, odd, or neither.
Solution: Recall that a function is even means f(x)=f(x) for all x and that a function is odd means f(x)=f(x) for all x. Recall that (x)3=(x)(x)(x)=x3. First calculate f(x)=7(x)3+4(x)1=7x34x1. Now calculate f(x)=(7x3+4x1)=7x34x+1. Since we see that f(x) is not the same as either f(x) (the signs of the first two terms don't match) or f(x) (the sign of the third term does not match), we conclude that it is neither even nor odd.