Odd and even functions
A function, f(x) is even if f(x) = f(-x). If f(x) = -f(x) then it is called an odd function.
f(x) = cos x is an even function.
f(x) = sin x is an odd function.
The functions x2, x4, x6 .. are even functions.
The functions x, x3, x5 .. are odd functions.
Question
A function f is defined by \(f(x)=arcsin\frac{x^2-1}{x^2+1},\;x\in \mathbb{R}\)
(a) Show that f is an even function. [1]
(b) By considering limits, show that the graph of y = f (x) has a horizontal asymptote and state its equation. [2]
(c) (i) Show that \(f{}'(x)=\frac{2x}{\sqrt{x^2}(x^2+1)} \; for\; x\in \mathbb{R},\;x\neq 0\)
(ii) By using the expression for f ′(x) and the result \(\sqrt{x^2}=\left | x \right |\) show that f is decreasing for x < 0 . [9]
A function g is defined by \(g(x)=arcsin\frac{x^2-1}{x^2+1},\;x\in \mathbb{R},\;x\geq 0.\)
(d) Find an expression for g-1(x) , justifying your answer. [5]
(e) State the domain of g-1 . [1]
(f) Sketch the graph of y = g-1(x) , clearly indicating any asymptotes with their equations and stating the values of any axes intercepts. [3]
▶️Answer/Explanation
Ans:
(a) Since $$\begin{eqnarray} f\left(-x\right) &=& \arcsin\left[\frac{\left(-x\right)^2-1}{\left(-x\right)^2+1}\right] \nonumber \\ &=& \arcsin\left(\frac{x^2-1}{x^2+1}\right) \nonumber \\ &=& f\left(x\right), \nonumber \\ \end{eqnarray}$$ $f$ is an even function.
(b) Consider $\lim_{x\to\infty} \arcsin\left(\frac{x^2-1}{x^2+1}\right)$, we have $$\begin{eqnarray} \lim_{x\to\infty} \arcsin\left(\frac{x^2-1}{x^2+1}\right) &=& \lim_{x\to\infty} \arcsin\left(\frac{1-\frac{1}{x^2}}{1+\frac{1}{x^2}}\right) \nonumber \\ &=& \arcsin\left(1\right) \nonumber \\ &=& \frac{\pi}{2}, \end{eqnarray}$$ i.e., $y=\frac{\pi}{2}$ is a horizontal asymptote.
(c)(i) Taking sine on both sides of $\arcsin\left(\frac{x^2-1}{x^2+1}\right)$, we have $\sin\left[f\left(x\right)\right]=\frac{x^2-1}{x^2+1}$.
Differentiating both sides with respect to $x$, we have $$\begin{eqnarray} f’\left(x\right)\cos\left[f\left(x\right)\right]=\frac{2x\left(x^2+1\right)-2x\left(x^2-1\right)}{\left(x^2+1\right)^2}. \end{eqnarray}$$ Rearranging, we have $$\begin{eqnarray} f’\left(x\right) &=& \frac{4x}{\left(x^2+1\right)^2}\times\frac{1}{\cos\left[f\left(x\right)\right]} \nonumber \\ &=& \frac{4x}{\left(x^2+1\right)^2}\times\frac{x^2+1}{2\sqrt{x^2}} \nonumber \\ &=& \frac{2x}{\sqrt{x^2}\left(x^2+1\right)}, \end{eqnarray}$$ since by Pythagoras’ Theorem, we can find $\cos\left[f\left(x\right)\right]=\frac{2\sqrt{x^2}}{x^2+1}$.
(c)(ii) When $x\lt 0$, $\sqrt{x^2}=\left|x\right|\gt 0$ and $x^2+1\gt 0$. Thus, $$\begin{eqnarray} \frac{2x}{\sqrt{x^2}\left(x^2+1\right)} \lt 0, \end{eqnarray}$$ i.e., $f$ is decreasing for $x\lt 0$.
(d) Let $y=\arcsin\left(\frac{x^2-1}{x^2+1}\right)$, then making $x$ the subject, we have $$\begin{eqnarray} \sin y=\frac{x^2-1}{x^2+1} \nonumber \\ \sin y\left(x^2+1\right) = x^2-1 \nonumber \\ x^2\left(\sin y-1\right) = -1-\sin y \nonumber \\ x^2 = \frac{1+\sin y}{1-\sin y} \nonumber \\ x = \pm\sqrt{\frac{1+\sin y}{1-\sin y}}. \end{eqnarray}$$ Since $x\geq 0$, $g^{-1}\left(x\right)=\sqrt{\frac{1+\sin x}{1-\sin x}}$.
(e) $\text{D}_{g^{-1}}=\text{R}_g=\left[-\frac{\pi}{2},\frac{\pi}{2}\right)$.
(f) (graph to be added in later)