Even and Odd Functions: Problems with Solutions

Solution:
Since [tex]f[/tex] is odd, [tex]f(-x)=-f(x)[/tex] for every real [tex]x[/tex]. The formula [tex]x^4[/tex] is valid only for [tex]x\ge 0[/tex], so we use it at [tex]x=1[/tex]:
[tex]f(1)=1^4=1\Rightarrow f(-1)=-f(1)=-1[/tex]
(Substituting [tex]x=-1[/tex] directly into [tex]x^4[/tex] would wrongly give [tex]1[/tex].)

Solution:
A function [tex]f[/tex] is even if its domain is symmetric with respect to [tex]0[/tex] and [tex]f(-x)=f(x)[/tex] for every [tex]x[/tex] in the domain.
Since [tex]x^2+1\gt 0[/tex], the domain is [tex]\mathbb{R}[/tex], which is symmetric. Using [tex]\sin (-x)=-\sin x[/tex], we get
[tex]f(-x)=\sqrt{(-x)^2+1}\,(-\sin x)^2=\sqrt{x^2+1}\,\sin ^2 x=f(x)[/tex]
Hence the function [tex]f[/tex] is even.

Solution:
A function [tex]g[/tex] is even if its domain is symmetric with respect to [tex]0[/tex] and [tex]g(-x)=g(x)[/tex] for every [tex]x[/tex] in the domain.
The domain consists of all [tex]x\ne \frac{\pi}{2}+k\pi[/tex], [tex]k\in\mathbb{Z}[/tex]. It is symmetric: if [tex]-x=\frac{\pi}{2}+k\pi[/tex], then [tex]x=\frac{\pi}{2}+(-k-1)\pi[/tex], so [tex]x[/tex] and [tex]-x[/tex] are excluded together. Using [tex]\tan (-x)=-\tan x[/tex], we get
[tex]g(-x)=(-x)^2-(-\tan x)^2=x^2-\tan ^2 x=g(x)[/tex]
Hence the function [tex]g[/tex] is even.

Solution:
A function [tex]h[/tex] is even if its domain is symmetric with respect to [tex]0[/tex] and [tex]h(-x)=h(x)[/tex] for every [tex]x[/tex] in the domain.
Both [tex]\tan x[/tex] and [tex]\cot x[/tex] must exist, i.e. [tex]x\ne\frac{k\pi}{2}[/tex], [tex]k\in\mathbb{Z}[/tex]. For such [tex]x[/tex] the denominator is not zero:
[tex]\tan x+\cot x=\frac{\sin ^2 x+\cos^2 x}{\sin x\cos x}=\frac{1}{\sin x\cos x}\ne 0[/tex]
So the domain is [tex]x\ne\frac{k\pi}{2}[/tex], [tex]k\in\mathbb{Z}[/tex], which is symmetric with respect to [tex]0[/tex]. Using [tex]\sin (-x)=-\sin x[/tex], [tex]\tan (-x)=-\tan x[/tex] and [tex]\cot (-x)=-\cot x[/tex], we get
[tex]h(-x)=\frac{-\sin x}{-\tan x-\cot x}=\frac{\sin x}{\tan x+\cot x}=h(x)[/tex]
Hence the function [tex]h[/tex] is even.

Solution:
The graph is symmetric with respect to the origin exactly when the function is odd: its domain is symmetric with respect to [tex]0[/tex] and [tex]f(-x)=-f(x)[/tex] for every [tex]x[/tex] in the domain.
For [tex]f(x)=\frac{2(x-1)}{x+1}[/tex] the domain [tex]x\ne -1[/tex] is not symmetric ([tex]1[/tex] belongs to it, [tex]-1[/tex] does not), so the function is not odd.
For [tex]f(x)=x^3-x^2-x-1[/tex] we have [tex]f(1)=-2[/tex] and [tex]f(-1)=-2[/tex], so [tex]f(-1)\ne -f(1)[/tex] and the function is not odd.
For [tex]f(x)=2^{1-x^2}[/tex] we have [tex]f(-x)=f(x)[/tex] and [tex]f(0)=2\ne 0[/tex], so the function is even, not odd (an odd function defined at [tex]0[/tex] satisfies [tex]f(0)=0[/tex]).
For [tex]f(x)=x\sqrt{1-x^2}[/tex] the domain [tex]\left[-1 , 1\right][/tex] is symmetric and
[tex]f(-x)=-x\sqrt{1-(-x)^2}=-x\sqrt{1-x^2}=-f(x)[/tex]
So this function is odd, and its graph is symmetric with respect to the origin.

Solution:
The domain is [tex]\mathbb{R}[/tex], which is symmetric. We have
[tex]f(-x)=2^{(a-2)x^2-(a-1)x-1}[/tex]
Since [tex]2^t[/tex] is strictly increasing, two of its values are equal only when the exponents are equal. Hence [tex]f(-x)=f(x)[/tex] for every [tex]x[/tex] is equivalent to
[tex](a-2)x^2-(a-1)x-1=(a-2)x^2+(a-1)x-1\iff 2(a-1)x=0[/tex]
for every real [tex]x[/tex]. Taking [tex]x=1[/tex] gives [tex]a=1[/tex]. Conversely, for [tex]a=1[/tex] we get [tex]f(x)=2^{-x^2-1}[/tex], which is even. Hence [tex]a=1[/tex].

Solution:
For [tex]x\ge 0[/tex] we have
[tex]f(x)=g(x)=x(x+4)(x-2)(x+2)[/tex]
and since [tex]x+4\gt 0[/tex] and [tex]x+2\gt 0[/tex] there, [tex]f(x)=0[/tex] only for [tex]x=0[/tex] and [tex]x=2[/tex].
For [tex]x\lt 0[/tex] we have [tex]-x\gt 0[/tex] and, since [tex]f[/tex] is odd, [tex]f(x)=-f(-x)[/tex]. Hence
[tex]f(x)=0\iff f(-x)=0\iff -x=2\iff x=-2[/tex]
The zeros [tex]-4[/tex] and [tex]-2[/tex] of the formula for [tex]g[/tex] cannot be taken automatically, because this formula gives [tex]f[/tex] only for [tex]x\ge 0[/tex]. For example,
[tex]f(-4)=-f(4)=-g(4)=-4\times 8\times 12=-384\ne 0[/tex]
So the solutions are [tex]-2[/tex], [tex]0[/tex] and [tex]2[/tex], i.e. the equation has [tex]3[/tex] solutions.

Solution:
Let [tex]x_0[/tex] be the unique root, so [tex]x_0[/tex] belongs to the domain and [tex]f(x_0)=5[/tex]. The domain of an even function is symmetric with respect to [tex]0[/tex], so [tex]-x_0[/tex] also belongs to it, and by evenness
[tex]f(-x_0)=f(x_0)=5[/tex]
Thus [tex]-x_0[/tex] is also a root of the equation. Since the root is unique,
[tex]-x_0=x_0\Rightarrow 2x_0=0\Rightarrow x_0=0[/tex]

Solution:
The domain of [tex]f[/tex] is [tex]x\ne -c[/tex], and the domain of an odd function is symmetric with respect to [tex]0[/tex]. If [tex]c\ne 0[/tex], then [tex]x=c[/tex] belongs to the domain (because [tex]c\ne -c[/tex]) but [tex]-c[/tex] does not, a contradiction. Hence [tex]c=0[/tex] and
[tex]f(x)=\frac{ax^2+b}{x}=ax+\frac{b}{x}[/tex]
which is odd for all [tex]a[/tex] and [tex]b[/tex]. From [tex]f(1)=a+b=2[/tex] we get [tex]b=2-a[/tex]. For [tex]x\gt 0[/tex], multiplying by [tex]x[/tex], the condition [tex]f(x)\ge 2[/tex] becomes [tex]p(x)\ge 0[/tex] for every [tex]x\gt 0[/tex], where
[tex]p(x)=ax^2-2x+2-a=(x-1)\left(ax-(2-a)\right)[/tex]
If [tex]a=0[/tex], then [tex]p(x)=-2(x-1)\lt 0[/tex] for [tex]x\gt 1[/tex], which is impossible.
If [tex]a\ne 0[/tex] and the second root [tex]\frac{2-a}{a}\ne 1[/tex], then [tex]x=1[/tex] is a simple root of [tex]p[/tex], so [tex]p[/tex] changes sign at [tex]x=1[/tex] and is negative at some positive [tex]x[/tex] close to [tex]1[/tex], which is impossible. Hence
[tex]\frac{2-a}{a}=1\Rightarrow a=1\Rightarrow b=2-a=1[/tex]
Check: for [tex]a=b=1[/tex], [tex]c=0[/tex] the function [tex]f(x)=\frac{x^2+1}{x}=x+\frac{1}{x}[/tex] is odd, [tex]f(1)=2[/tex], and for [tex]x\gt 0[/tex]
[tex]f(x)-2=\frac{(x-1)^2}{x}\ge 0[/tex]
So the only such function is [tex]f(x)=\frac{x^2+1}{x}[/tex].