(int_(" with "f'(y)=(((sqrt(y+6))-1)/(3)))(((sqrt(y+6))-1)/(3)))

Consider f: R _(+) to [-5, oo) given by f (x) = 9x ^(2) + 6x -5. Show that f is invertible with f ^(-1) (y) = ((sqrt(y +6) -1)/(3)).

Consider f: R _(+) to [-5, oo) given by f (x) = 9x ^(2) + 6x -5. Show that f is invertible with f ^(-1) (y) = ((sqrt(y +6) -1)/(3)).

Consider f: R_+->[-5,oo) given by f(x)=9x^2+6x-5 . Show that f is invertible with f^(-1)(y)=(((sqrt(y+6))-1)/3)

Consider f:R rarr[-5,oo) given by f(x)=9x^(2)+6x-5. show that f is invertible with f^(-1)(y)=((sqrt(y+6)-1)/(3))

The value of f(x,y)=((x^(3)y)^((1)/(4))-(y^(3)x)^((1)/(4)))/(sqrt(y)-sqrt(x))+(1+sqrt(xy))/((xy)^((1)/(4)))xx((1+2sqrt((y)/(x))+(y)/(x))^(4) when x=9 and y=0.04

The tangent to the graph of the function y=f(x) at the point with abscissae x=1, x=2, x=3 make angles pi/6,pi/3 and pi/4 respectively. The value of int_1^3f\'(x)f\'\'(x)dx+int_2^3f\'\'(x)dx is (A) (4-3sqrt(3))/3 (B) (4sqrt(3)-1)/(3sqrt(3)) (C) (4-3sqrt(3))/2 (D) (3sqrt(3)-1)/2

Consider f : R_(+) rarr [-5,oo) given by f(x) = 9x^(2) +6x-5 . Show that f is invertible with f^(-1)(y) = ((sqrt(y+6)-1)/3)

A square of side a lies above the X- axis and has one vertex at the origin . The side passing through the origin makes an angle pi//6 with the positive direction of X-axis .The equation of its diagonal not passing through the origin is y(sqrt(3)-1)-x(1-sqrt(3))=2a y(sqrt(3)+1) +x(1-sqrt(3))=2a y(sqrt(3)+1)+x(1+sqrt(3)) =2a y(sqrt(3)+1)+x(sqrt(3)-1)=2a

Consider f:RR_(+) rarr [-5, infty) given by f(x)=9x^(2)+6x-5 . Show that f is invertible with f^(-1) (y) =(sqrt(y+6)-1)/(3)