Q.If $f(x)=x^3-\dfrac{1}{x^3},\ x\neq 0$, then show that $f(x)+f\left(\dfrac{1}{x}\right)=0$.

$f(-x)$ turns out to be the reciprocal of $f(x)$.

Given $f(x)=\dfrac{1-x}{1+x}$, replace $x$ by $-x$:

$f(-x)=\frac{1-(-x)}{1+(-x)}=\frac{1+x}{1-x}.$

Now compare with $f(x)=\dfrac{1-x}{1+x}$. Notice

$f(-x)=\frac{1+x}{1-x}=\frac{1}{\dfrac{1-x}{1+x}}=\frac{1}{f(x)}.$

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The range of $\sin^{-1}x$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.

The function $y=\sin x$ is not one-to-one over all of $\mathbb{R}$, so its inverse is defined by restricting the angle to the principal branch $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, on which $\sin$ is one-to-one and takes every value in $[-1,1]$.

Hence:

• Domain of $\sin^{-1}x$: $[-1,1]$ …

Substituting $x+1$ into $f(x)=x^2-x+1$ gives $f(x+1)=x^2+x+1$, so the answer is option (c).

Given $f(x) = x^2 - x + 1$, replace every $x$ with $(x+1)$:

$f(x+1) = (x+1)^2 - (x+1) + 1.$

Expand:

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Because $0<2$, use the branch $f(x)=x+2$, giving $f(0)=2$.

The function is defined piecewise:

$f(x) = \begin{cases} x^2 - 4x & \text{if } x \ge 2 \\ x + 2 & \text{if } x < 2 \end{cases}$

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Check $f(-x)$ against $f(x)$ and $-f(x)$; only $x^3+5$ matches neither, so it is neither even nor odd.

Recall: even $\Rightarrow f(-x)=f(x)$; odd $\Rightarrow f(-x)=-f(x)$.

• (a) $f(x)=x^{10}$: $f(-x)=x^{10}=f(x)$ — even.
• (c) $f(x)=x^{2}$: $f(-x)=x^{2}=f(x)$ — even.
• (d) $f(x)=x^{5}$: $f(-x)=-x^{5}=-f(x)$ — odd. …

Substituting $-x$ gives $f(-x)=\dfrac{1+x}{1-x}$, exactly the reciprocal of $f(x)$, so $f(-x)=\dfrac{1}{f(x)}$.

Given:

$f(x) = \frac{1-x}{1+x}.$

Replace $x$ with $-x$:

$f(-x) = \frac{1-(-x)}{1+(-x)} = \frac{1+x}{1-x}.$

Now compare with $\dfrac{1}{f(x)}$: …