What is the equation of the line tangent to #f(x)= sqrt(3x^3-2x) # at #x=2#?

Question

What is the equation of the line tangent to #f(x)=  sqrt(3x^3-2x)   # at #x=2#?

Evelyn Agard · Accepted Answer

Tangent line: #y=(17x-34+6sqrt(10))/(3sqrt(2))#

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.

Ezra Adams · Answer

undefined To find the equation of the line tangent to the function f(x) = sqrt(3x^3 - 2x) at x = 2, we need to find the derivative of the function and evaluate it at x = 2. The derivative of f(x) is given by f'(x) = (9x^2 - 2) / (2 * sqrt(3x^3 - 2x)). Evaluating f'(x) at x = 2, we get f'(2) = (9(2)^2 - 2) / (2 * sqrt(3(2)^3 - 2(2))). Simplifying this expression, we have f'(2) = (36 - 2) / (2 * sqrt(24 - 4)). Further simplifying, f'(2) = 34 / (2 * sqrt(20)). Finally, simplifying the expression, we get f'(2) = 17 / sqrt(20). Therefore, the equation of the line tangent to f(x) at x = 2 is y = f(2) + f'(2)(x - 2), where f(2) represents the value of f(x) at x = 2.