Stationary Points?

Question

Two questions I'm having trouble with
1) Find the stationary points on the curve y=(3x-1)(x-2)^4
2) Show that f(x)=1/x^3 has no stationary points

Answer 1

ok

1.

when dy/dx is = 0 you have the "maxima and minima" or "stationary" points



so differentiate to get with the product rule to get



dy/dx = 4(3x - 1)(x - 2)^3 + 3(x - 2)^4



= (x - 2)^3(12x - 1)



let this = 0



so (x - 2)^3 = 0 or 12x - 1 = 0



so x = 2 or x = 1/12



put back into original equation y=(3x-1)(x-2)^4 to find y



so stationary points are (2,0) and (0.083,-7.03)



2.

x^-3 = y



dy/dx = -3x^-2 = 0

-3/(x^2) = 0



-3 = 0 which isn't true therefore it has no stationary points

Answer 2

1) differentiate it to find the derivative by using dy/dx. then equate the derivative to zero and solve for x. substitute these x values back into the original equation and solve for y. these will give you your stationary point(s)



2) same thing. differentiate it. find the derivative. equate the derivative for zero. solve for x. if x cannot be solved (maybe because you may have to find the square root of a negative number, for example) then there are no stationary points.

Answer 3

y=(3x-1)(x-2)^4

y'=3*4(x-2)^3

dy/dx=12(x-2)^3

u have equalate this eq to 0 to find critical point

12(x-2)^3=0

x=2;

this is crical point

so stationary point is (x,f(x))

=(2,0)



ii)f(x)=1/x^3

f'=-3/x^4

0=-3/x^4

x=0;

since critical point is zero,it has no stationary point

Answer 4

1) You need to take the dy/dx here is a link for you to look over. When dy/dx=zero that is a stationary point.

2) plug in fx for x and then take dy/dx and show that for dy/dx for (-inf,inf)<>zero

Answer 5

i suggest u go to a tutor and get help with this answer as now one is that smart

Answer 6

Sorry don't know. = )