Is \(g=\{(1,1),(2,3),(3,5),(4,7)\}\) a function? Justify. If this is described by the relation \(g(x)=\alpha x+\beta\), then what values should be assigned to \(\alpha\) and \(\beta\)?
Yes. A relation is a function if each input (first coordinate) is associated with exactly one output (second coordinate). Here the first coordinates \(1,2,3,4\) are distinct and each appears exactly once, so \(g\) is a function.
Assume \(g(x)=\alpha x+\beta\). Use any two ordered pairs to determine \(\alpha,\beta\). From \(g(1)=1\) we get \(\alpha(1)+\beta=1\Rightarrow \alpha+\beta=1\). From \(g(2)=3\) we get \(2\alpha+\beta=3\). Subtract the first from the second: \(\alpha=2\). Then \(\beta=1-\alpha=1-2=-1\).
Hence \(\alpha=2,\;\beta=-1\).
Find the domain of each of the following functions:
(i) \(f(x)=\dfrac{1}{\sqrt{1-\cos x}}\)
(ii) \(f(x)=\dfrac{1}{\sqrt{x+|x|}}\)
(iii) \(f(x)=x|x|\)
(iv) \(f(x)=\dfrac{x^3-x+3}{x^2-1}\)
(v) \(f(x)=\dfrac{3x}{2x-8}\)
(i) Require denominator \(\sqrt{1-\cos x}\) to be defined and nonzero. So \(1-\cos x>0\Rightarrow\cos x\ne1\). Therefore \(x\ne 2n\pi\) for integer \(n\). Domain: \(\mathbb{R}\setminus\{2n\pi\mid n\in\mathbb{Z}\}\).
(ii) Inside square root: \(x+|x|\ge0\) and denominator nonzero. Note \(x+|x|=\begin{cases}2x,&x\ge0\\0,&x<0\end{cases}\). For \(x<0\) the expression is \(0\) giving division by zero, so excluded. For \(x\ge0\), \(x+|x|=2x\ge0\) and positive for \(x>0\). Hence domain: \(x>0\). That is the positive reals \((0,\infty)\) (often denoted \(\mathbb{R}^+\)).
(iii) The expression \(x|x|\) is defined for all real \(x\) (absolute value always defined). Domain: \(\mathbb{R}\).
(iv) Rational function with denominator \(x^2-1\). Exclude zeros of denominator: \(x^2-1=0\Rightarrow x=\pm1\). Domain: \(\mathbb{R}\setminus\{-1,1\}\).
(v) Rational function with denominator \(2x-8\). Exclude \(2x-8=0\Rightarrow x=4\). Domain: \(\mathbb{R}\setminus\{4\}\).
Find the range of the following functions:
(i) \(f(x)=\dfrac{3}{2-x^2}\)
(ii) \(f(x)=1-|x-2|\)
(iii) \(f(x)=|x-3|\)
(iv) \(f(x)=1+3\cos2x\)
(i) Let \(y=\dfrac{3}{2-x^2}\). Rearranging gives \(x^2=2-\dfrac{3}{y}\). For real \(x\) we need \(\dfrac{2y-3}{y}\ge0\). This inequality holds for \(y\ge\tfrac{3}{2}\) or \(y<0\). Also \(y\ne0\). Thus the range is \((-\infty,0)\cup[3/2,\infty)\).
(ii) Since \(|x-2|\ge0\), maximum of \(1-|x-2|\) is \(1\) (when \(|x-2|=0\)). As \(|x-2|\to\infty\) the expression tends to \(-\infty\). Hence range: \(( -\infty,1]\).
(iii) Absolute value yields nonnegative values. So range: \([0,\infty)\).
(iv) As \(\cos2x\in[-1,1]\), we have \(1+3\cos2x\in[1-3,\;1+3]=[-2,4]\). Hence range: \([-2,4]\).
Redefine the function \(f(x)=|x-2|+|2+x|\) for \(-3\le x\le3\) as a piecewise function (i.e. give \(f(x)\) on intervals determined by the absolute-value expressions).
We partition \([-3,3]\) by the points where expressions inside absolute values change sign: \(x=-2\) and \(x=2\). Evaluate on intervals:
For \(-3\le x<-2\): \(x-2<0\) and \(2+x<0\). So \(|x-2|=-(x-2)=-x+2\) and \(|2+x|=-(2+x)=-x-2\). Thus \(f(x)=(-x+2)+(-x-2)=-2x\).
For \(-2\le x<2\): \(x-2<0\) but \(2+x\ge0\). So \(|x-2|=-x+2,\;|2+x|=x+2\). Hence \(f(x)=(-x+2)+(x+2)=4\) (constant).
For \(2\le x\le3\): both \(x-2\ge0\) and \(2+x\ge0\). So \(|x-2|=x-2,\;|2+x|=x+2\). Then \(f(x)=(x-2)+(x+2)=2x\).
Therefore the piecewise definition on \([-3,3]\) is:
\(f(x)=\begin{cases}-2x,&-3\le x<-2\\4,&-2\le x<2\\2x,&2\le x\le3.\end{cases}\)
If \(f(x)=\dfrac{x-1}{x+1}\), then show that
(i) \(f\bigl(\dfrac{1}{x}\bigr)=-f(x)\)
(ii) \(f\bigl(-\dfrac{1}{x}\bigr)=-\dfrac{1}{f(x)}\)
(i) Compute \(f\bigl(\dfrac{1}{x}\bigr)=\dfrac{\tfrac{1}{x}-1}{\tfrac{1}{x}+1}=\dfrac{1-x}{1+x}\cdot\dfrac{1}{x}\div\dfrac{1}{x}=\dfrac{1-x}{1+x}\). Simplify: \(\dfrac{1-x}{1+x}=-\dfrac{x-1}{x+1}=-f(x)\). Thus \(f(1/x)=-f(x)\).
(ii) Compute \(f\bigl(-\dfrac{1}{x}\bigr)=\dfrac{-\tfrac{1}{x}-1}{-\tfrac{1}{x}+1}=\dfrac{-(1+ x)}{-(1-x)}\) after multiplying numerator and denominator by \(x\). Simplify: \(\dfrac{-(1+x)}{-(1-x)}=\dfrac{1+x}{1-x}\). On the other hand \(f(x)=\dfrac{x-1}{x+1}\) so \(\dfrac{1}{f(x)}=\dfrac{x+1}{x-1}\) and hence \(-\dfrac{1}{f(x)}=-\dfrac{x+1}{x-1}=\dfrac{1+x}{1-x}\). Therefore \(f(-1/x)=-\dfrac{1}{f(x)}\).
Let \(f(x)=\sqrt{x}\) and \(g(x)=x\) be two functions defined in the domain \(\mathbb{R}^+\cup\{0\}\). Find
(i) \((f+g)(x)\)
(ii) \((f-g)(x)\)
(iii) \((fg)(x)\)
(iv) \(\bigl(\dfrac{f}{g}\bigr)(x)\)
Domain for both: \([0,\infty)\).
(i) \((f+g)(x)=f(x)+g(x)=\sqrt{x}+x\).
(ii) \((f-g)(x)=\sqrt{x}-x\).
(iii) \((fg)(x)=f(x)g(x)=x\sqrt{x}=x^{3/2}\).
(iv) \(\bigl(\dfrac{f}{g}\bigr)(x)=\dfrac{\sqrt{x}}{x}=\dfrac{1}{\sqrt{x}}\) for \(x>0\). (Note: at \(x=0\) this quotient is not defined.)
Find the domain and range of the function \(f(x)=\dfrac{1}{\sqrt{x-5}}\).
Inside the square root we need \(x-5>0\) (strictly positive because denominator cannot be zero). Hence domain: \(x>5\) i.e. \((5,\infty)\).
For \(x>5\) we have \(\sqrt{x-5}>0\) so \(f(x)>0\). As \(x\to5^+\), \(\sqrt{x-5}\to0^+\) so \(f(x)\to+\infty\). As \(x\to\infty\), \(\sqrt{x-5}\to\infty\) so \(f(x)\to0^+\). Thus range is \((0,\infty)\) (all positive real numbers).
If \(f(x)=y=\dfrac{ax-b}{cx-a}\), then prove that \(f(y)=x\).
We are given \(y=\dfrac{ax-b}{cx-a}\). We solve this relation for \(x\) in terms of \(y\) and then show that replacing the variable by \(y\) in the formula for \(f\) returns \(x\).
Start with \(y=\dfrac{ax-b}{cx-a}\). Cross-multiply:
\(y(cx-a)=ax-b\Rightarrow cxy-ay=ax-b\).
Collect terms in \(x\): \(cxy-ax = ay-b\Rightarrow x(cy-a)=ay-b\).
Provided \(cy-a\ne0\), we obtain
\(x=\dfrac{ay-b}{cy-a}\).
But the right-hand side is exactly the same functional form as \(f\) with \(x\) replaced by \(y\):
\(f(y)=\dfrac{ay-b}{cy-a}=x\).
Therefore \(f(y)=x\), as required (under the nondegeneracy condition that denominators are nonzero).