# Answer in Calculus for Sarita bartwal #209278

Show that the function f: R^2→R^2 given by

f(x,y) = (xy^3+1, x^2+y^2) is not invertible. Futher check whether it is locally invertible at the point (2,1)

Given function f: R^{2} “to” R^{2}

such that f(x,y) = (xy^{3}+1, x^{2}+y^{2})

we have to show that f is not invertible in R^{2}

Let u = xy^{3}+1,v = x^{2}+y^{2}

Df(x,y) = “begin{vmatrix}n frac{delta u}{delta x} & frac{delta u}{delta y} \n frac{delta v}{delta x} & frac{delta v}{delta y}nend{vmatrix}”

Df(x,y) = “begin{vmatrix}n frac{delta }{delta x}(xy^3+1) & frac{delta }{delta y}(xy^3+1) \n frac{delta }{delta x}(x^2+y^2) & frac{delta }{delta y}(x^2+y^2)nend{vmatrix}”

Df(x,y) = “begin{vmatrix}n y^3 & 3xy^2 \n 2x & 2ynend{vmatrix}”

Df(x,y) = 2y^{4}“-“ 6x^{2}y^{2}

Df(x,y) = 2y^{2}(y^{2}“-“ 3x^{2})

here we see that, Df(x,y) = 0 if 2y^{2}(y^{2}“-“ 3x^{2}) = 0 “implies” either y = 0 or (y^{2}“-“ 3x^{2}) = 0 “implies” either y = 0 or y = “pm” “sqrt{3}x”

hence,

** Df(x,y) is not invertble at (a,0) **“in”** R**^{2}** for any a **“in”** R also Df(x,y) is not invertble at (x, **“pmsqrt{3}x”**) **“in”** R**^{2}

“implies”** f(x,y) is not invertible.**

at point (2,1)

Df(2,1) = 2(1)^{2}[(1)^{2} “-“ 3(2)^{2}]

Df(2,1) = 2(1 “-“ 12)

Df(2,1) = -22 “neq” 0

hence, **f(x,y) is locally invertible at (2,1)**