Questions
Question 1
- For each function \(f\) and domain \(D\), sketch \(f\) on its restriction to \(D\).
- \(f(x)=4x+1\), \(D=[-3,1)\)
- \(f(x)=-2x+1\), \(D=(-\infty,3)\)
- \(f(x)=x^2\), \(D=[-2,-1]\cup [0,1)\cup(2,4)\)
- \(f(x)=\sqrt{x}\), \(D=[1,4]\cup [9,16]\)
- \(f(x)=x^3\), \(D=(-2,0)\cup [1,3]\)
- \(f(x)=1\), \(D=[-3,-1]\cup (0,1]\cup [3,\infty)\)
Question 2
- For each function \(f\) and \(g\), compute \(f+g, f\cdot g, f-g,\frac{f}{g}, f\circ g\), and \(g\circ f\).
- \(f(x)=x-1\), \(g(x)=\sqrt{x}\)
- \(f(x)=3x\), \(g(x)=x^2\)
- \(f(x)=x^2+x-1\), \(g(x)=\frac{1}{x}\)
- \(f(x)=\frac{x-1}{x+2}\) or \(g(x)=\sqrt{x-1}\)
Question 3
- For each function \(f\) and \(g\), determine the domain of \(f+g, f\cdot g, f-g,\frac{f}{g},\frac{g}{f}\)
- functions \(f\) and \(g\) with \(\mathcal{D}(f)=[-3,10)\), \(\mathcal{D}(g)=[-2,14)\), zero set of \(f\) is \(\{-2,2,4\}\) and zero set of \(g\) is \(\{1,11\}\).
- functions \(f\) and \(g\) with \(\mathcal{D}(f)=(-\infty,0]\cup[1,5)\), \(\mathcal{D}(g)=[-1,\infty)\), zero set of \(f\) is \(\{1\}\) and zero set of \(g\) is \(\{10\}\).
- functions \(f\) and \(g\) with \(\mathcal{D}(f)=(-\infty,\infty)\), \(\mathcal{D}(g)=[-10,10)\), zero set of \(f\) is \(\{-1,[0,1],2\}\) and zero set of \(g\) is \(\{-5,5\}.\)
Question 4
- Take \[a(x)=x,\quad b(x)=x^2,\quad c(x)=1,\quad d(x)=\sqrt{x},\quad e(x)=\frac{1}{x},\quad\text{and}\quad f(x)=2x.\] Compute each of the function \(g.\)
- \(g=a+b+\frac{c+d}{a}\)
- \(g=e\circ d+b\cdot f+b\circ f+f\circ b\)
- \(g=d\circ (e+c)+a-e\)
- \(g=e\circ e\circ f\)
- \(g=e\circ(b+a+c)+d\circ (b+a+c)\)
Question 5
- Take \[a(x)=x,\quad b(x)=x^2,\quad c(x)=1,\quad d(x)=\sqrt{x},\quad e(x)=\frac{1}{x},\quad\text{and}\quad f(x)=2x.\] Decompose \(g\) into sums, products, and composites of the functions \(a,b,c,d,e\) and \(f\).
- \(g(x)=x+\sqrt{x}\)
- \(g(x)=x^2\sqrt{x-1}\)
- \(g(x)=\frac{x+1}{\sqrt{2x}}\)
- \(g(x)=\frac{x^2-2}{2x}+2x^2+x\sqrt{x}\)
Question 6
- For each function \(f\), determine the range of \(f\).
- \(f(x)=\frac{2}{x-1}\)
- \(f(x)=\frac{2x+4}{x-1}\)
- \(f(x)=\sqrt{x}+1\)
Answers
Question 2
- \[(f+g)(x)=x-1+\sqrt{x}\] \[(f\cdot g)(x)=(x-1)\cdot\sqrt{x}=x\sqrt{x}-\sqrt{x}\] \[(f-g)(x)=x-1-\sqrt{x}\] \[\left(\frac{f}{g}\right)(x)=\frac{x-1}{\sqrt{x}}\] \[\left(f\circ g\right)(x)=\sqrt{x}-1\] \[\left(g\circ f\right)(x)=\sqrt{x-1}\]
- \[(f+g)(x)=3x+x^2\] \[(f\cdot g)(x)=3x^3\] \[(f-g)(x)=3x-x^2\] \[\left(\frac{f}{g}\right)(x)=\frac{3x}{x^2}=\text{ or }\frac{3}{x}\] \[\left(f\circ g\right)(x)=3x^2\] \[\left(g\circ f\right)(x)=9x^2\]
- \[(f+g)(x)=x^2+x-1+\frac{1}{x}\] \[(f\cdot g)(x)=(x^2+x-1)\frac{1}{x}\text{ or }x+1-\frac{1}{x}\] \[(f-g)(x)=x^2+x-1-\frac{1}{x}\] \[\left(\frac{f}{g}\right)(x)=\frac{x^2+x-1}{\frac{1}{x}}\text{ or }x^3+x^2-x\] \[\left(f\circ g\right)(x)=\frac{1}{x^2}+\frac{1}{x}-1\] \[\left(g\circ f\right)(x)=\frac{1}{x^2+x-1}\]
- \[(f+g)(x)=\frac{x-1}{x+2}+\sqrt{x-1}\] \[(f\cdot g)(x)=\frac{x-1}{x+2}\cdot \sqrt{x-1}\] \[(f-g)(x)=\frac{x-1}{x+2}-\sqrt{x-1}\] \[\left(\frac{f}{g}\right)(x)=\frac{\frac{x-1}{x+2}}{\sqrt{x-1}}\text{ or }\frac{x-1}{(x+2)\sqrt{x-1}}\] \[\left(f\circ g\right)(x)=\frac{\sqrt{x-1}-1}{\sqrt{x-1}+2}\] \[\left(g\circ f\right)(x)=\sqrt{\frac{x-1}{x+2}-1} \text{ or } \sqrt{-\frac{3}{x+2}}\]
Question 3
- Domain of \(f+g\), \(fg\) and \(f-g\) is \([-2,10)\). Domain of \(\frac{f}{g}\) is \([-2,1)\cup(1,10)\). Domain of \(\frac{g}{f}\) is \((-2,2)\cup(2,4)\cup(4,10)\).
- Domain of \(f+g\), \(fg\) and \(f-g\) is \([-1,0]\cup[1,5)\). Domain of \(\frac{f}{g}\) is \([-1,0]\cup[1,5)\). Domain of \(\frac{g}{f}\) is \([-1,0]\cup(1,5)\)
- Domain of \(f+g\), \(fg\) and \(f-g\) is \([-10,10)\). Domain of \(\frac{f}{g}\) is \([-10,-5)\cup(-5,5)\cup(5,10)\). Domain of \(\frac{g}{f}\) is \([-10,-1)\cup(-1,0)\cup(1,2)\cup(2,10)\)
Question 4
- \(x+x^2+\frac{1+\sqrt{x}}{x}\)
- \(\frac{1}{\sqrt{x}}+2x^3+4x^2+2x^2\).
- \(\sqrt{\frac{1}{x}+1}+x-\frac{1}{x}\)
- \(\frac{1}{\frac{1}{2x}}\) or \(2x\)
- \(\frac{1}{x^2+x+1}+\sqrt{x^2+x+1}\)
Question 5
- \(g=a+d\)
- \(g=b\cdot d\circ(a-c)\)
- \(g=\frac{a+c}{d\circ f}\)
- \(g=\frac{b-c-c}{f}+f\circ b+a\cdot d\)
Question 6
- \(\mathcal{R}(f)=(-\infty,0)\cup(0,\infty)\)
- \(\mathcal{R}(f)=(-\infty,2)\cup(2,\infty)\)
- \(\mathcal{R}(f)=[1,\infty)\)