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Question Number 143684 by Huy last updated on 17/Jun/21

x^3 +x−1=^3 (√(2x^3 +11))+(√(5x^2 +16))  Find x∈R

$$\mathrm{x}^{\mathrm{3}} +\mathrm{x}−\mathrm{1}=^{\mathrm{3}} \sqrt{\mathrm{2x}^{\mathrm{3}} +\mathrm{11}}+\sqrt{\mathrm{5x}^{\mathrm{2}} +\mathrm{16}} \\ $$$$\mathrm{Find}\:\mathrm{x}\in\mathbb{R} \\ $$

Answered by TheHoneyCat last updated on 17/Jun/21

let x_0 =2  x_0  is a solution (2^3 +2−1=9=^3 (√(27))+(√(36)))    Juging by the graph, it seems it is the only solution  let me proove it  let x∈]x_0 ,+∞[  let f: x→x^3 +x−1  and g: x→^3 (√(2x^3 +11))+(√(5x^2 +16))    (df/dx)=3x^2 +1  (dg/dx)=6x^2 (2x^3 +11)^(−(2/3)) +10x(5x^2 +16)^(−(1/2))     x≥2  ⇒2x^3 ≥16  ⇒2x^3 +11≥27  ⇒(2x^3 +11)^(1/3) ≥3  ⇒(2x^3 +11)^(2/3) ≥9  ⇒(2x^3 +11)^((−2)/3) ≤(1/9)  ⇒6x^2 (2x^3 +11)^((−2)/3) ≤(2/9)3x^2     5x^2 +16≥5x^2   ⇒(5x^2 +16)^(1/2) ≥(√5)x  ⇒10x(5x^2 +16)^((−1)/2) ≤((10)/( (√5)))    and 3x^2 +1≥(2/9)3x^2 +((10)/( (√5)))      so (df/dx)≥(dg/dx)   so ∀x≥x_0  f(x)≠g(x)    the same reasonning can be donne for  x∈[0,x_0 ]  and for x<0 checking the sine is sufficient_■

$$\mathrm{let}\:{x}_{\mathrm{0}} =\mathrm{2} \\ $$$${x}_{\mathrm{0}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}\:\left(\mathrm{2}^{\mathrm{3}} +\mathrm{2}−\mathrm{1}=\mathrm{9}=^{\mathrm{3}} \sqrt{\mathrm{27}}+\sqrt{\mathrm{36}}\right) \\ $$$$ \\ $$$$\mathrm{Juging}\:\mathrm{by}\:\mathrm{the}\:\mathrm{graph},\:\mathrm{it}\:\mathrm{seems}\:\mathrm{it}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution} \\ $$$$\mathrm{let}\:\mathrm{me}\:\mathrm{proove}\:\mathrm{it} \\ $$$$\left.\mathrm{let}\:{x}\in\right]{x}_{\mathrm{0}} ,+\infty\left[\right. \\ $$$$\mathrm{let}\:{f}:\:\mathrm{x}\rightarrow\mathrm{x}^{\mathrm{3}} +\mathrm{x}−\mathrm{1} \\ $$$$\mathrm{and}\:{g}:\:\mathrm{x}\rightarrow^{\mathrm{3}} \sqrt{\mathrm{2x}^{\mathrm{3}} +\mathrm{11}}+\sqrt{\mathrm{5x}^{\mathrm{2}} +\mathrm{16}} \\ $$$$ \\ $$$$\frac{{df}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\frac{{dg}}{{dx}}=\mathrm{6}{x}^{\mathrm{2}} \left(\mathrm{2}{x}^{\mathrm{3}} +\mathrm{11}\right)^{−\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{10}{x}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{16}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$ \\ $$$${x}\geqslant\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{3}} \geqslant\mathrm{16} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{3}} +\mathrm{11}\geqslant\mathrm{27} \\ $$$$\Rightarrow\left(\mathrm{2}{x}^{\mathrm{3}} +\mathrm{11}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \geqslant\mathrm{3} \\ $$$$\Rightarrow\left(\mathrm{2}{x}^{\mathrm{3}} +\mathrm{11}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \geqslant\mathrm{9} \\ $$$$\Rightarrow\left(\mathrm{2}{x}^{\mathrm{3}} +\mathrm{11}\right)^{\frac{−\mathrm{2}}{\mathrm{3}}} \leqslant\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\Rightarrow\mathrm{6}{x}^{\mathrm{2}} \left(\mathrm{2}{x}^{\mathrm{3}} +\mathrm{11}\right)^{\frac{−\mathrm{2}}{\mathrm{3}}} \leqslant\frac{\mathrm{2}}{\mathrm{9}}\mathrm{3}{x}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{16}\geqslant\mathrm{5}{x}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{16}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \geqslant\sqrt{\mathrm{5}}{x} \\ $$$$\Rightarrow\mathrm{10}{x}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{16}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \leqslant\frac{\mathrm{10}}{\:\sqrt{\mathrm{5}}} \\ $$$$ \\ $$$$\mathrm{and}\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}\geqslant\frac{\mathrm{2}}{\mathrm{9}}\mathrm{3}{x}^{\mathrm{2}} +\frac{\mathrm{10}}{\:\sqrt{\mathrm{5}}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{so}\:\frac{{df}}{{dx}}\geqslant\frac{{dg}}{{dx}}\: \\ $$$$\mathrm{so}\:\forall{x}\geqslant{x}_{\mathrm{0}} \:{f}\left({x}\right)\neq{g}\left({x}\right) \\ $$$$ \\ $$$${the}\:{same}\:{reasonning}\:{can}\:{be}\:{donne}\:{for} \\ $$$${x}\in\left[\mathrm{0},{x}_{\mathrm{0}} \right] \\ $$$$\mathrm{and}\:\mathrm{for}\:{x}<\mathrm{0}\:\mathrm{checking}\:\mathrm{the}\:\mathrm{sine}\:\mathrm{is}\:\mathrm{sufficient}_{\blacksquare} \\ $$

Commented by TheHoneyCat last updated on 17/Jun/21

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