At #x=2#
#color(white)("XXX")f(2)=sqrt(3*2^3-2*2)=sqrt(24-4)=2sqrt(5)#
So we are looking for the tangent at the #color(red)("point "(2,2sqrt(5)))#
The general slope of the tangent is given by the derivative of the function.
By letting #color(blue)(g(x)=3x^3-2x)#
we can rewrite the given function as
#color(white)("XXX")color(green)(f(color(red)(g(x)))=color(red)(g(x))^(1/2)#
Then, using the Chain Rule, we have the general slope:
#color(white)("XXX")(df(g(x)))/(dx)=color(cyan)((df(g(x)))/(dg(x))) * color(magenta)((dg(x))/(dx))#
#color(white)("XXXXXXXXX")=color(cyan)(1/2(3x^3-2x)^(-1/2)) * color(magenta)(9x^2-2)#
#color(white)("XXXXXXXXX")=(9x^2-2)/(2sqrt(3x^3-2x))#
When #x=2#, this general slope becomes
#color(white)("XXX")color(green)(m)=(9*2^2-2)/(2sqrt(3*2^3-2*2))=34/(6sqrt(2))=color(green)(17/(3sqrt(2)))#
With the point #color(red)((x_0,y_0)=(2,2sqrt(5)))# and a slope of #color(green)(m=17/(3sqrt(2)))#
we can write the equation of the tangent using the point-slope form:
#color(white)("XXX")(y-color(red)(y_0))/(x-color(red)(x_0))=color(green)(m)#
#color(white)("XXX")(y-color(red)(2sqrt(5)))/(x-color(red)(2))=color(green)(17/(3sqrt(2))#
#color(white)("XXX")3sqrt(2)y-6sqrt(10)=17x-34#
#color(white)("XXX")y=(17x-34+6sqrt(10))/(3sqrt(2))#
...you could play with this a bit, but I do not see that the result is going to become any nicer looking than this.
