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Section 2.1 #36: For the function
f(x)={3,x≤−2x2+6,x>−2,
find the values f(−5),f(−2),f(0), and f(2).
Solution: Since −5≤−2, we compute
f(−5)=3.
Since −2≤−2, 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,x≥0.
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)=√2−x,
a.) Find the domain of f, g, f+g, f−g, fg, ff, fg, and gf.
b.) Find (f+g)(x), (f−g)(x), (fg)(x), (ff)(x), (fg)(x), and (gf)(x).
Solution:
a.) First note that the domain of f is defined by the inequality x≥0, or as an interval,
dom(f)=[0,∞).
The domain of g is defined by the inequality 2−x≥0, which is equivalent to 2≥x. We can write 2≥x as x≤2, or as an interval,
dom(g)=(−∞,2].
The domain of f+g, f−g, and fg are all just the intersection of the domain of f with the domain of g:
dom(f+g)=dom(f−g)=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+√2−x,
(f−g)(x)=f(x)−g(x)=√x−√2−x,
(fg)(x)=f(x)g(x)=√x√2−x,
(ff)(x)=f(x)f(x)=√x√x=x,
(fg)(x)=f(x)g(x)=√x√2−x,
and
(gf)(x)=g(x)f(x)=√2−x√x.
Section 2.2 #50: Compute the difference quotient f(x+h)−f(x)h, where f(x)=4x−1.
Solution: Compute directly:
f(x+h)−f(x)h=[4(x+h)−1]−[4x−1]h=4x+4h−1−4x+1h=4hh=4.