Question Number 214251 by Hanuda354 last updated on 02/Dec/24
Commented by Hanuda354 last updated on 02/Dec/24
$$\mathrm{Determine}\:\:\mathrm{where}\:\:{f}\:\:\mathrm{is}\:\:\mathrm{continuous}\:\:\mathrm{algebraically}. \\ $$$$\mathrm{Write}\:\:\mathrm{in}\:\:\mathrm{interval}\:\:\mathrm{notation}. \\ $$
Answered by a.lgnaoui last updated on 03/Dec/24
![1•]−∞,0[ f(x)=log(5x+20) valable pour 5x+20>0 x>−4 so Range of= D_f =]−4,+∞[ ∩ ]−∞,0[=]−4,0[ limf(x)_(x→−4^+ ) =+∞ f(0=2log2+log5 f′(x)=(1/(x+4)) continue in ]−4,0[ 2• sur Interval :]0,5] ,f(x)=((3(x−1)(x−(2/3)))/((x+11)(x−((24)/(11))))) valable pour x#{−((24)/(11)),11 } donc continue sur ]0,5] les bornes limf(x)_(x→0^+ ) =f(0)=−(1/(12)) = limf(x)_(x→5^− ) =((4×(13))/(−6×79))=((−572)/(474)) or lim f(x)_(x→0^− ) ≠limf(x) _(x→0^+ ) donc f(x) n est pas continue au point(0) 3•f(x)=3x^2 −13x−8 Df=]5,+∞[ limf(x)_(x→5^+ ) =2 (f non continue au pt(5) limf(x)_(x→5^(−1) ) ≠limf(x)_(x→5^+ )](https://www.tinkutara.com/question/Q214257.png)
$$\left.\mathrm{1}\bullet\right]−\infty,\mathrm{0}\left[\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\boldsymbol{\mathrm{log}}\left(\mathrm{5}\boldsymbol{\mathrm{x}}+\mathrm{20}\right)\right. \\ $$$$\:\:\boldsymbol{\mathrm{valable}}\:\boldsymbol{\mathrm{pour}}\:\mathrm{5}\boldsymbol{\mathrm{x}}+\mathrm{20}>\mathrm{0}\:\:\:\boldsymbol{\mathrm{x}}>−\mathrm{4} \\ $$$$\left.\boldsymbol{\mathrm{so}}\:\:\mathrm{Range}\:\mathrm{of}=\:\boldsymbol{\mathrm{D}}_{\mathrm{f}} =\right]−\mathrm{4},+\infty\left[\:\cap\:\right]−\infty,\mathrm{0}\left[=\right]−\mathrm{4},\mathrm{0}\left[\right. \\ $$$$\:\:\boldsymbol{\mathrm{limf}}\left(\boldsymbol{\mathrm{x}}\right)_{\boldsymbol{\mathrm{x}}\rightarrow−\mathrm{4}^{+} } =+\infty\:\:\:\mathrm{f}\left(\mathrm{0}=\mathrm{2log2}+\mathrm{log5}\right. \\ $$$$\left.\:\:\:\mathrm{f}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}+\mathrm{4}}\:\mathrm{continue}\:\mathrm{in}\:\right]−\mathrm{4},\mathrm{0}\left[\right. \\ $$$$\left.\mathrm{2}\left.\bullet\:\mathrm{sur}\:\mathrm{Interval}\:\:\::\right]\mathrm{0},\mathrm{5}\right]\:\:\: \\ $$$$,\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\frac{\mathrm{3}\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{2}}{\mathrm{3}}\right)}{\left(\boldsymbol{\mathrm{x}}+\mathrm{11}\right)\left(\boldsymbol{\mathrm{x}}−\frac{\mathrm{24}}{\mathrm{11}}\right)}\:\:\boldsymbol{\mathrm{val}}\mathrm{a}\boldsymbol{\mathrm{ble}}\:\boldsymbol{\mathrm{pou}}\mathrm{r}\:\boldsymbol{\mathrm{x}}#\left\{−\frac{\mathrm{24}}{\mathrm{11}},\mathrm{11}\:\right\}\: \\ $$$$\left.\:\left.\:\:\:\:\:\mathrm{donc}\:\:\mathrm{continue}\:\mathrm{sur}\:\:\:\right]\mathrm{0},\mathrm{5}\right] \\ $$$$\mathrm{les}\:\mathrm{bornes} \\ $$$$\mathrm{limf}\left(\mathrm{x}\right)_{\mathrm{x}\rightarrow\mathrm{0}^{+} } =\mathrm{f}\left(\mathrm{0}\right)=−\frac{\mathrm{1}}{\mathrm{12}}\:= \\ $$$$\boldsymbol{\mathrm{limf}}\left(\boldsymbol{\mathrm{x}}\right)_{\mathrm{x}\rightarrow\mathrm{5}^{−} } =\frac{\mathrm{4}×\left(\mathrm{13}\right)}{−\mathrm{6}×\mathrm{79}}=\frac{−\mathrm{572}}{\mathrm{474}} \\ $$$$\mathrm{or}\:\boldsymbol{\mathrm{lim}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}^{−} } \neq\mathrm{limf}\left(\mathrm{x}\right)\:_{\mathrm{x}\rightarrow\mathrm{0}^{+} } \\ $$$$\mathrm{donc}\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\:\:\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{est}}\:\boldsymbol{\mathrm{pas}}\:\boldsymbol{\mathrm{continue}}\:\boldsymbol{\mathrm{au}}\:\boldsymbol{\mathrm{point}}\left(\mathrm{0}\right) \\ $$$$\mathrm{3}\bullet\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{13}\boldsymbol{\mathrm{x}}−\mathrm{8} \\ $$$$\left.\:\:\:\boldsymbol{\mathrm{Df}}=\right]\mathrm{5},+\infty\left[\right. \\ $$$$\:\:\boldsymbol{\mathrm{limf}}\left(\boldsymbol{\mathrm{x}}\right)_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{5}^{+} } =\mathrm{2}\:\:\:\left(\boldsymbol{\mathrm{f}}\:\boldsymbol{\mathrm{non}}\:\boldsymbol{\mathrm{continue}}\:\boldsymbol{\mathrm{au}}\:\boldsymbol{\mathrm{pt}}\left(\mathrm{5}\right)\right. \\ $$$$\mathrm{limf}\left(\mathrm{x}\right)_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{5}^{−\mathrm{1}} } \neq\mathrm{limf}\left(\mathrm{x}\right)_{\mathrm{x}\rightarrow\mathrm{5}^{+} } \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 03/Dec/24
Commented by a.lgnaoui last updated on 03/Dec/24
$$\mathrm{allure}\:\mathrm{de}\:\mathrm{courbes}\:\::\:\mathrm{f}\left(\mathrm{x}\right)\cup\mathrm{g}\left(\mathrm{x}\right)\cup\mathrm{h}\left(\mathrm{x}\right) \\ $$