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Section 2.1 #36: For the function f(x)={3,x2x2+6,x>2, find the values f(5),f(2),f(0), and f(2).
Solution: Since 52, we compute f(5)=3. Since 22, we compute f(2)=3. Since 0>2, we compute f(0)=02+6=0+6=6. Since 2>2, we compute f(2)=22+6=1+6=7.

Section 2.1 #39: Graph the following function: f(x)={x2,x<0x+3,x0. Solution: First consider the graph of y=x2:

Now consider the graph of y=x+3:

Therefore the graph of the piecewise function is


Section 2.2 #24: For the functions f(x)=x and g(x)=2x,
a.) Find the domain of f, g, f+g, fg, fg, ff, fg, and gf.
b.) Find (f+g)(x), (fg)(x), (fg)(x), (ff)(x), (fg)(x), and (gf)(x).
Solution:
a.) First note that the domain of f is defined by the inequality x0, or as an interval, dom(f)=[0,).
The domain of g is defined by the inequality 2x0, which is equivalent to 2x. We can write 2x as x2, or as an interval, dom(g)=(,2].
The domain of f+g, fg, and fg are all just the intersection of the domain of f with the domain of g: dom(f+g)=dom(fg)=dom(fg)=[0,)(,2]=[0,2]. The domain of (ff)(x) will simply be the domain of f: dom(ff)=dom(f)=[0,). The domain of (fg)(x) is the intersection of the domains of f and g and also taking away the x-values that cause g(x) to be zero: dom(fg)=[0,)(,2]andtakeaway2=[0,2). The domain of (gf)(x) is the intersection of the domains of f and g and also taking away the x-values that cause f(x) to be zero: dom(gf)=[0,)(,2]andtakeaway0=(0,2]. b.) Compute (f+g)(x)=f(x)+g(x)=x+2x, (fg)(x)=f(x)g(x)=x2x, (fg)(x)=f(x)g(x)=x2x, (ff)(x)=f(x)f(x)=xx=x, (fg)(x)=f(x)g(x)=x2x, and (gf)(x)=g(x)f(x)=2xx.

Section 2.2 #50: Compute the difference quotient f(x+h)f(x)h, where f(x)=4x1.
Solution: Compute directly: f(x+h)f(x)h=[4(x+h)1][4x1]h=4x+4h14x+1h=4hh=4.