Question Number 83977 by john santu last updated on 08/Mar/20
$$\mathrm{find}\:\mathrm{range}\:\mathrm{function}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{x}\sqrt{\mathrm{7x}−\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{without} \\ $$$$\mathrm{calculus} \\ $$
Answered by john santu last updated on 08/Mar/20
Answered by TANMAY PANACEA last updated on 08/Mar/20
![x×(√(−x^2 +7x−1)) x×(√(−1−(x^2 −2×x×(7/2)+((49)/4)−((49)/4)))) x×(√(−1+((49)/4)−(x−(7/2))^2 )) x×(√(((45)/4)−(x−(7/2))^2 )) (x−(7/2))^2 ≯((45)/4) ((45)/4)−(x−(7/2))^2 ≥0 (((3(√5))/2)+x−(7/2))(((3(√5))/2)−x+(7/2))≥0 critical value of x=((7−3(√5))/2) and ((7+3(√5))/2) (x−a)(b−x)≥0 [a=((7−3(√5))/2) b=((7+3(√( 5)))/2) b>a] g(x)=(x−a)(b−x) when x>b g(x)<0 when x<a g(x)<0 when b>x>a g(x)>0 g(x)=0 when x=a and x=b f(x)=x×(√(−x^2 +7x−1)) f(x)=x×(√((x−a)(b−x))) [b>a] f(a)=f(b)=0 but x≯b, x≮a wait...](https://www.tinkutara.com/question/Q83988.png)
$${x}×\sqrt{−{x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{1}}\: \\ $$$${x}×\sqrt{−\mathrm{1}−\left({x}^{\mathrm{2}} −\mathrm{2}×{x}×\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{49}}{\mathrm{4}}−\frac{\mathrm{49}}{\mathrm{4}}\right)}\: \\ $$$${x}×\sqrt{−\mathrm{1}+\frac{\mathrm{49}}{\mathrm{4}}−\left({x}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${x}×\sqrt{\frac{\mathrm{45}}{\mathrm{4}}−\left({x}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} }\: \\ $$$$\left({x}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} \ngtr\frac{\mathrm{45}}{\mathrm{4}} \\ $$$$\frac{\mathrm{45}}{\mathrm{4}}−\left({x}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\left(\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}+{x}−\frac{\mathrm{7}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}−{x}+\frac{\mathrm{7}}{\mathrm{2}}\right)\geqslant\mathrm{0} \\ $$$${critical}\:{value}\:{of}\:{x}=\frac{\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\:{and}\:\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\left({x}−{a}\right)\left({b}−{x}\right)\geqslant\mathrm{0}\:\:\:\left[{a}=\frac{\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:{b}=\frac{\mathrm{7}+\mathrm{3}\sqrt{\:\mathrm{5}}}{\mathrm{2}}\:\:{b}>{a}\right] \\ $$$${g}\left({x}\right)=\left({x}−{a}\right)\left({b}−{x}\right) \\ $$$${when}\:{x}>{b}\:\:{g}\left({x}\right)<\mathrm{0} \\ $$$${when}\:{x}<{a}\:\:{g}\left({x}\right)<\mathrm{0} \\ $$$${when}\:\:\:{b}>{x}>{a}\:\:{g}\left({x}\right)>\mathrm{0} \\ $$$${g}\left({x}\right)=\mathrm{0}\:\:{when}\:{x}={a}\:\:\:{and}\:\:{x}={b} \\ $$$${f}\left({x}\right)={x}×\sqrt{−{x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{1}}\: \\ $$$${f}\left({x}\right)={x}×\sqrt{\left({x}−{a}\right)\left({b}−{x}\right)}\:\:\:\left[{b}>{a}\right] \\ $$$${f}\left({a}\right)={f}\left({b}\right)=\mathrm{0}\:\:\:\:{but}\:{x}\ngtr{b},\:\:{x}\nless{a} \\ $$$${wait}… \\ $$$$ \\ $$
Commented by john santu last updated on 08/Mar/20
$$\mathrm{domain}\:\frac{\mathrm{45}}{\mathrm{4}}\:−\left({x}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} \geqslant\:\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\:\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{7}}{\mathrm{2}}−\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\leqslant\:\mathrm{0} \\ $$$$\frac{\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\:\leqslant\:{x}\:\leqslant\:\frac{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by john santu last updated on 08/Mar/20
$$\mathrm{range}\:\mathrm{y}\:=\:\sqrt{−{x}^{\mathrm{4}} +\mathrm{7}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} } \\ $$
Commented by john santu last updated on 08/Mar/20
$$\mathrm{y}\:'\:=\:\frac{−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{21}{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{2}\sqrt{−{x}^{\mathrm{4}} +\mathrm{7}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} }}\:=\:\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{3}} −\mathrm{21}{x}^{\mathrm{2}} +\mathrm{2}{x}=\mathrm{0} \\ $$$${x}\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{21}{x}+\mathrm{2}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{21}}{\mathrm{4}}{x}+\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{21}}{\mathrm{8}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{441}}{\mathrm{64}}\:=\:\mathrm{0} \\ $$