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[j] = 0 and [-§] = —1. Hence, this function is neither even nor odd.
f(x)=
f(x)=
/« = |*-1|/(1) = 0 and /(-1) = |-1-1| = 2. So, fix) is neither even nor odd.
The function J(x) of Problem 5.11.
J(—x) = —(—x) \—x\ = x \x\ = —J(x), so this is an odd function.
fix) = 2x + l.
/(I) = 3 and /(-!)=-!. So, f(x) is neither even nor odd.
Show that a function f(x) is even if and only if its graph is symmetric with respect to the y-axis.
Assume that fix) is even. Let (jc, y) be on the graph of/. We must show that (—x, y) is also on the graph of
/. Since (x, y) is on the graph of /, f(x) = y. Hence, since/is even, f(~x)=f(x) = y, and, thus, (—x, y) is
on the graph of/. Conversely, assume that the graph of/is symmetric with respect to the _y-axis. Assume that
fix) is defined and f(x) = y. Then (x, y) is on the graph of/. By assumption, (-x, y) also is on the graph of/.
Hence, f(~x) = y. Then, /(-*)=/(*), and fix) is even.
Show that/(*) is odd if and only if the graph of/is symmetric with respect to the origin.
Assume that fix) is odd. Let (x, y) be on the graph of/. Then fix) = y. Since fix) is odd, fi—x) =
—fix) = —y, and, therefore, (—x, -y) is on the graph of /. But, (x, y) and (~x, -y) are symmetric with
respect to the origin. Conversely, assume the graph of / is symmetric with respect to the origin. Assume
fix) = y. Then, (x, y) is on the graph of/. Hence, by assumption, (—x, -y) is on the graph of/. Thus,
fi-x) = — y = -fix), and, therefore, fix) is odd.
Show that, if a graph is symmetric with respect to both the x-axis and the y-axis, then it is symmetric with respect
to the origin.
Assume (x, y) is on the graph. Since the graph is symmetric with respect to the jt-axis, (x, -y) is also on the
graph, and, therefore, since the graph is symmetric with respect to the y-axis, (—x, —y) is also on the graph.
Thus, the graph is symmetric with respect to the origin.
Show that the converse of Problem 5.50 is false.
The graph of the odd function fix) = x is symmetric with respect to the origin. However, (1,1) is on the
graph but (—1,1) is not; therefore, the graph is not symmetric with respect to the y-axis. It is also not symmetric
with respect to the x-axis.
If / is an odd function and /(O) is denned, must /(O) = 0?
Yes. /(0)=/(-0) = -/(0). Hence, /(0) = 0.
If fix) = x
2 + kx + 1 for all x and / is an even function, find k.
il) = 2 + k and /(-l) = 2-fc. By the evenness of/, /(-!) = /(!). Hence, 2+k = 2-k, k =
-k, k = 0.
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fix) = [x]
Hence, this function is odd.
So, fix) is odd.
[j] = 0 and [-§] = —1. Hence, this function is neither even nor odd.
f(x)=
f(x)=
/« = |*-1|/(1) = 0 and /(-1) = |-1-1| = 2. So, fix) is neither even nor odd.
The function J(x) of Problem 5.11.
J(—x) = —(—x) \—x\ = x \x\ = —J(x), so this is an odd function.
fix) = 2x + l.
/(I) = 3 and /(-!)=-!. So, f(x) is neither even nor odd.
Show that a function f(x) is even if and only if its graph is symmetric with respect to the y-axis.
Assume that fix) is even. Let (jc, y) be on the graph of/. We must show that (—x, y) is also on the graph of
/. Since (x, y) is on the graph of /, f(x) = y. Hence, since/is even, f(~x)=f(x) = y, and, thus, (—x, y) is
on the graph of/. Conversely, assume that the graph of/is symmetric with respect to the _y-axis. Assume that
fix) is defined and f(x) = y. Then (x, y) is on the graph of/. By assumption, (-x, y) also is on the graph of/.
Hence, f(~x) = y. Then, /(-*)=/(*), and fix) is even.
Show that/(*) is odd if and only if the graph of/is symmetric with respect to the origin.
Assume that fix) is odd. Let (x, y) be on the graph of/. Then fix) = y. Since fix) is odd, fi—x) =
—fix) = —y, and, therefore, (—x, -y) is on the graph of /. But, (x, y) and (~x, -y) are symmetric with
respect to the origin. Conversely, assume the graph of / is symmetric with respect to the origin. Assume
fix) = y. Then, (x, y) is on the graph of/. Hence, by assumption, (—x, -y) is on the graph of/. Thus,
fi-x) = — y = -fix), and, therefore, fix) is odd.
Show that, if a graph is symmetric with respect to both the x-axis and the y-axis, then it is symmetric with respect
to the origin.
Assume (x, y) is on the graph. Since the graph is symmetric with respect to the jt-axis, (x, -y) is also on the
graph, and, therefore, since the graph is symmetric with respect to the y-axis, (—x, —y) is also on the graph.
Thus, the graph is symmetric with respect to the origin.
Show that the converse of Problem 5.50 is false.
The graph of the odd function fix) = x is symmetric with respect to the origin. However, (1,1) is on the
graph but (—1,1) is not; therefore, the graph is not symmetric with respect to the y-axis. It is also not symmetric
with respect to the x-axis.
If / is an odd function and /(O) is denned, must /(O) = 0?
Yes. /(0)=/(-0) = -/(0). Hence, /(0) = 0.
If fix) = x
2 + kx + 1 for all x and / is an even function, find k.
il) = 2 + k and /(-l) = 2-fc. By the evenness of/, /(-!) = /(!). Hence, 2+k = 2-k, k =
-k, k = 0.
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fix) = [x]
Hence, this function is odd.
So, fix) is odd.