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Question Number 206328 by mnjuly1970 last updated on 12/Apr/24

$$$$$$f\left(x\right)=\lfloor \frac{1+x+x^2}{x+x^2}\rfloor isgiven.$$$$\Rightarrow \{\begin{array}{l}{D}_f=?\left(domain\right)\\ R_f=?\left(range\right)\end{array}$$$$$$

Answered by TheHoneyCat last updated on 12/Apr/24

$$D_{\lfloor \bullet \rfloor }=\mathbb{R}$$$$\mathrm{So}\mathrm{we}\mathrm{need}\frac{1+x+x^2}{x+x^2}=\frac{1}{x+x^2}+1\in \mathbb{R}$$$$\mathrm{So}\mathrm{we}\mathrm{need}\frac{1}{x}\times \frac{1}{1+x}\in \mathbb{R}$$$$\mathrm{This}\mathrm{works}\mathrm{on}\mathrm{all}\mathbb{R}\mathrm{\setminus }\{0,-1\}$$$$\mathrm{buy}\mathrm{we}\mathrm{could}\mathrm{also}\mathrm{add}\{+\mathrm{\infty },-\mathrm{\infty }\}$$$$\mathrm{and}\mathrm{also}\{x\in \mathbb{C}\mid \mathrm{Arg}\left(x\right)=-\mathrm{Arg}(1+x)\}$$$$\mathrm{This}\mathrm{boils}\mathrm{down}\mathrm{to}:\mid x\mid e^{i\theta }(1+\mid x\mid e^{i\theta })\in \mathbb{R}$$$$\Rightarrow e^{i\theta }+\mid x\mid e^{i2\theta }\in \mathbb{R}$$$$\Rightarrow \sin \theta +\mid x\mid \sin \left(2\theta \right)=0$$$$$$$$\mathrm{But}\mathrm{assuming}\mathrm{you}\mathrm{just}\mathrm{work}\mathrm{with}\mathrm{real}\mathrm{numbers}:$$$$D_f=\mathbb{R}\mathrm{\setminus }\{-1,0\}=]-\mathrm{\infty },-1[\cup ]-1,0[\cup ]0,\mathrm{\infty }[$$$$$$$$R_f=\lfloor R_{\frac{1}{x+x^2}+1}\rfloor$$$$=\lfloor R_{\frac{1}{x+x^2}}+1\rfloor$$$$=\lfloor {\left(R_{x+x^2\mid x\in \mathbb{R}\mathrm{\setminus }\{-1,0\}}\right)}^{-1}\rfloor +1$$$$\mathrm{Now},x+x^2\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{of}\mathrm{deg}2$$$$\mathrm{it}\mathrm{has}\mathrm{its}\mathrm{minimu}\mathrm{reached}\mathrm{at}\frac{-1}{2}$$$$\mathrm{ie}{R}_{x+x^2}=[\frac{-1}{4},+\mathrm{\infty }[$$$$\mathrm{so}:$$$$R_{x+x^2\mid x\in \mathbb{R}\mathrm{\setminus }\{-1,0\}}=[-1/4,0[\cup {\mathbb{R}}_{+}^{\ast }$$$$\mathrm{so}:$$$$({R_{x+x^2\mid x\in \mathbb{R}\mathrm{\setminus }\{-1,0\}})}^{-1}=]-\mathrm{\infty },-4]\cup {\mathbb{R}}_{+}^{\ast }$$$$\mathrm{Thus}:$$$$R_f=(\lfloor ]-\mathrm{\infty },-4]\rfloor +1)\cup (\lfloor {\mathbb{R}}_{+}^{\ast }\rfloor +1)$$$$=(\lfloor {\mathbb{R}}_{-}\rfloor -4+1)\cup ({\mathbb{Z}}_{+}+1)$$$$=\mid ]-\mathrm{\infty },-3\mid ]\cup [\mid 1,\mathrm{\infty }[\mid$$$$\{\begin{array}{l}{D}_f=\mathbb{R}\mathrm{\setminus }\{-1,0\}\\ R_f=\mathbb{Z}\mathrm{\setminus }\{-2,-1,0\}{}_{◻}\end{array}$$

Commented by mr W last updated on 12/Apr/24

$$forwhichxsuchthatf\left(x\right)=-3?$$

Commented by TheHoneyCat last updated on 12/Apr/24

$$f(-1/2)=\lfloor \frac{1}{(-1/2)+{(-1/2)}^2}+1\rfloor$$$$=\lfloor \frac{1}{-1/2+1/4}+1\rfloor$$$$=\lfloor \frac{1}{-1/4}+1\rfloor$$$$=\lfloor -4+1\rfloor$$$$=\lfloor -3\rfloor$$$$=-3$$$$\mathrm{But}\mathrm{since}\mathrm{we}\mathrm{are}\mathrm{talking}\mathrm{about}\mathrm{it},$$$$\mathrm{notice}\mathrm{that}\mathrm{this}\mathrm{point}\mathrm{is}\mathrm{very}\mathrm{special}.$$$$\mathrm{Most}\mathrm{points}x\mathrm{are}\mathrm{in}\mathrm{a}\mathrm{constant}\mathrm{neigboorhood}.$$$$\mathrm{meaning}x\pm ϵ\mathrm{with}\mathrm{epsilon}\mathrm{tiny}\mathrm{enough}$$$$\mathrm{will}\mathrm{give}\mathrm{the}\mathrm{same}\mathrm{value}.$$$$\mathrm{some}\mathrm{points}\mathrm{are}\mathrm{such}\mathrm{that}\mathrm{this}\mathrm{holds}\mathrm{only}$$$$\mathrm{for}x+ϵ\mathrm{or}x-ϵ.(\mathrm{points}\mathrm{where}\mathrm{the}\lfloor \bullet \rfloor \mathrm{causes}$$$$\mathrm{a}\mathrm{jump})$$$$\mathrm{But}{x}_0=-1/2\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{point}$$$$\mathrm{that}\mathrm{has}\mathrm{a}\mathrm{compleet}\mathrm{neigboorhood}\mathrm{that}\mathrm{is}\mathrm{worth}$$$$-4\mathrm{and}\mathrm{never}\mathrm{itself}.$$$$$$$$\mathrm{this}\mathrm{means}{\lim }_{x\to x_0}f\left(x\right)=-4\ne f\left(x_0\right)$$$$$$$$\mathrm{In}\mathrm{more}\mathrm{fancy}\mathrm{words},f\mathrm{is}\mathrm{continuous}\mathrm{by}\mathrm{part}$$$$\mathrm{everywhere}\mathrm{except}\mathrm{at}{x}_0.$$

Commented by mr W last updated on 12/Apr/24

$$great!$$

Answered by MM42 last updated on 12/Apr/24

$$f\left(x\right)=\left[\frac{1}{x^2+x}\right]+1$$$$ifg\left(x\right)=\frac{1}{x^2+x}\Rightarrow f\left(x\right)=\left[g\left(x\right)\right]+1$$$$D_f=D_g=\mathbb{R}-\{0,1\}$$$$x\to \{\begin{array}{l}-1^{-}\Rightarrow g\to +\mathrm{\infty }\\ -1^{+}\Rightarrow g\to -\mathrm{\infty }\end{array}\mathrm{\&}x\to \{\begin{array}{l}{0}^{-}\Rightarrow g\to -\mathrm{\infty }\\ 0^{+}\Rightarrow g\to +\mathrm{\infty }\end{array}$$$$g^{\prime }\left(x\right)=\frac{-2x-1}{{(x^2+x)}^2}\Rightarrow M(-\frac{1}{2},-4)$$$$\Rightarrow R_g=(-\mathrm{\infty },-4]\cup (0,\mathrm{\infty })$$$$R_f=\mathbb{Z}-\{-3,-2,-1\}$$$$$$

Commented by MM42 last updated on 12/Apr/24

Commented by TheHoneyCat last updated on 14/Apr/24

Hey, sorry to be "mr bad news" here, but I think both your D and R are false... i think you are missing a minus sign in D (no big deal, you probably just went too fast). But more problematically, I believe your R is wrong. You must not remove "-3". This is visible on your own drawing, but also on the discussion Mr. W and I had previously (check f(-1/2) and you'll see).

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