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Section 2.3 #23: Find (f∘g)(x) and (g∘f)(x) and the domain of each:
f(x)=41−5x;g(x)=1x.
Solution: First compute
(f∘g)(x)=f(g(x))=f(1x)=41−5x=4xx−5.
The domain of (f∘g) 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 (f∘g) is all real numbers except 0 and 5.
Now compute
(g∘f)(x)=g(f(x))=g(41−5x)=141−5x=1−5x4.
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. 41−5x=5. Multiply by 1−5x to get 4=5−25x implying −25x=−1, or x=125. Therefore the domain of (g∘f) is all real numbers except for 15 and 125.
Section 2.3 #40: Find f(x) and g(x) such that h(x)=(f∘g)(x), where
h(x)=3√x2−8.
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 x2−8. So we can take f(x)=3√x and g(x)=x2−8 so that
(f∘g)(x)=f(g(x))=f(x2−8)=3√x2−8.
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+4x−1 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=−7x3−4x−1.
Now calculate
−f(x)=−(7x3+4x−1)=−7x3−4x+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.