Question

Calculate the coordinates of the point on the curve y=(1-x)ˆ4 at which the gradient is -4.

Answer 1

You will need to expand that equation, find dy/dx then set dy/dx equal to -4. Solving this equation for x should give you a set of x-coordinates that you can then plug back into your original equation to find the corresponding y-coordinates.

Answer 2

I expanded the equation out but still couldn't get the x..I got a decimal

Answer 3

y = x^4 - 4x^3 + 6x^2 - 4x +1

Answer 4

you got this... yeah? ^^^^^^

Answer 5

and deriviative:

dy/dx = 4x^2 - 12x^2 + 12x - 4

Answer 6

I got -xˆ3+3xˆ2-3x

Answer 7

and as derivative is the formula of the gradient at any point in time
so:
when dy/dx = -4
-4 = 4x^2 - 12x^2 + 12x - 4

Answer 8

so it only works at x=0

Answer 9

Isn't dy/dx just -4(1-x)ˆ3

Answer 10

so gradient = -4 at x = 0
you can sub x=0 into your initial eqn, or if ur really bright you can see tht the corresponding y value would equal c

Answer 11

"Isn't dy/dx just -4(1-x)ˆ3 "
no
dy/dx is actually 4 (x-1)^3

Answer 12

if u do it longhand and expand u cant make those mistakes easily though

Answer 13

ooo, shiny, cheers ;)

Answer 14

No but it's a -x

Answer 15

So 4x(-1x1)=-4

Answer 16

yeah, but you're not doing the chain or product rule correctly... but i cant really help with that, tht was @.Sam. and @campbell_st that were doing it that way

Answer 17

sorry, before derivative is :
dy/dx = 4x^3 - 12x^2 + 12x - 4

i accidently put:
dy/dx = 4x^2 - 12x^2 + 12x - 4

Answer 18

Is it?
\[ y=(1-x)^4\]

Answer 19

yep

Answer 20

Chain rule?
\[y'=4(1-x)^3 \cdot\frac{d}{dx}[1-x]\]

Answer 21

You get
\[y'=-4(1-x)^3\]

Answer 22

weird... i expanded and got something completely different

Answer 23

... u sure @.Sam. ...?
i reduced mine down and got dy/dx is actually 4 (x-1)^3

Answer 24

Mine is different, yours is 4(x-1)^3 mine is -4(1-x)^3

Answer 25

Is just playing with the equations

Answer 26

ah... cool
cheers

Answer 27

cheers

Answer 28

got to love them odd powers:)