How to find the coordinates of the stationary points on the curve #y = x^3 – 6x^2 – 36x + 16#?

Question

Julia Adams · Accepted Answer

We have a local maximum at #(-2,56)# and a local minimum at #(6,-200)# and an inflexion point at #(2,-72)#

We start with the first derivative

#y=x^3-6x^2-36x+16#

#dy/dx=3x^2-12x-36#

The critical points are when #dy/dx=0#

That is,

#3x^2-12x-36=0#

#2(x^2-4x-12)=0#

#2(x+2)(x-6)=0#

Therefore,

#x=-2# and #x=6#

We build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaa)##6##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+2##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-6##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##dy/dx##color(white)(aaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##y##color(white)(aaaaaaaaa)##↗##color(white)(aaaa)##↘##color(white)(aaaa)##↗#

Now, we calculate the second derivative

#(d^2y)/dx^2=6x-12#

We have an inflexion point when, #(d^2y)/dx^2=0#

That is, #x=2#

We make a second chart

#color(white)(aaaa)##Interval##color(white)(aaaa)##]-oo,2[##color(white)(aaaa)##]2,+oo[#

#color(white)(aaaa)##(d^2y)/dx^2##color(white)(aaaaaaaaaa)##-##color(white)(aaaaaaaa)##+#

#color(white)(aaaa)##y##color(white)(aaaaaaaaaaaaa)##nn##color(white)(aaaaaaaa)##uu#

We have a local maximum at #(-2,56)# and a local minimum at #(6,-200)# and an inflexion point at #(2,-72)#

Jameson Allen · Answer

undefined To find the coordinates of the stationary points on the curve $ y = x^3 - 6x^2 - 36x + 16 $, first, find the derivative of the function. Then, set the derivative equal to zero and solve for $ x $. Finally, plug the values of $ x $ into the original function to find the corresponding $ y $ coordinates.