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Question-91166




Question Number 91166 by student work last updated on 28/Apr/20
Commented by Prithwish Sen 1 last updated on 28/Apr/20
the curves  y=x^2  and y=e^x  can meet at atmost two points.^
$$\mathrm{the}\:\mathrm{curves} \\ $$$$\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \:\mathrm{can}\:\mathrm{meet}\:\mathrm{at}\:\mathrm{atmost}\:\mathrm{two}\:\mathrm{points}\overset{} {.} \\ $$
Commented by jagoll last updated on 28/Apr/20
using Lambert W function
$${using}\:{Lambert}\:{W}\:{function} \\ $$
Commented by 675480065 last updated on 28/Apr/20
true  but can newton raphson not work
$$\mathrm{true} \\ $$$$\mathrm{but}\:\mathrm{can}\:\mathrm{newton}\:\mathrm{raphson}\:\mathrm{not}\:\mathrm{work} \\ $$
Commented by mr W last updated on 28/Apr/20
it asks only the number of solutions,  not the solutions self.
$${it}\:{asks}\:{only}\:{the}\:{number}\:{of}\:{solutions}, \\ $$$${not}\:{the}\:{solutions}\:{self}. \\ $$
Commented by Prithwish Sen 1 last updated on 28/Apr/20
  x ⋍ −0.7034    (approx.)
$$\:\:\boldsymbol{\mathrm{x}}\:\backsimeq\:−\mathrm{0}.\mathrm{7034}\:\:\:\:\left(\boldsymbol{\mathrm{approx}}.\right) \\ $$
Commented by mr W last updated on 28/Apr/20
there is only one solution which is  x=−2 W((1/2))≈−0.7034
$${there}\:{is}\:{only}\:{one}\:{solution}\:{which}\:{is} \\ $$$${x}=−\mathrm{2}\:\mathbb{W}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\approx−\mathrm{0}.\mathrm{7034} \\ $$
Commented by MJS last updated on 28/Apr/20
there′s no other solution ∈C
$$\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{other}\:\mathrm{solution}\:\in\mathbb{C} \\ $$

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