algebra in finding stationary pts of y = x + 4/(x+1)

[Iamadam]

Hey all

I had a question on my maths paper today, which I had NO idea how to answer.
It was to find the stationary points of:

y = x + 4/(x+1)

So I go ahead and differentiate:

y = x + 4(x+1)^-1
y`= 1 - 4(x+1)^-2
y`= 1 - 4/((x+1)^-2)

And equate it to 0....

1 - 4/((x+1)^-2) = 0

And then I'm stuck! I can't believe I got stuck at something which seems to be trivial algebra. Help would be appreciated - I would like to know how to approach a question like this in future (since I'm sure I should know how to do this!)

Thanks a lot.

-Adam

 

[arthur ohlsten]

y=x+ 4/[x+1]
y=x +4[x+1]^-1
take derivative
dy/dx =1 -4[x+1]^-2
set dy/dx=0
4/[x+1]^2=1
4=[x+1]^2
4=x^2+2x+1
x^2+2x-3=0
factor
[x+3][x-1]=0
x=1 or x=-3 answer

if you plot the curve:
at x=-oo y=-oo oo is infinity
as x increases y decreases ,reachs a maximum at x=-3
as x increases y decreases and is assymtotic to x=-1 ,y=-oo
as x increases y decreases from y=+00 at x=-1+
at x=0 y=4
as x increases y decreases is a minimum at x=1
as x increase y increase is assymtotic with a slope of 1 and y=oo at x=oo

Arthur

 

[stapel]

y = x + 4(x+1)^-1
y`= 1 - 4(x+1)^-2
y`= 1 - 4/((x+1)^-2)

When you move the base to the other side of the fraction line, the sign on the power changes. So:

. . . . .y' = 1 - 4(x + 1)^-2

...becomes:

. . . . .y' = 1 - 4/(x + 1)²

Setting equal to zero, one obtains:

. . . . .1 - 4/(x + 1)² = 0

. . . . .1 = 4/(x + 1)²

"Cross multiply" to get:

. . . . .1(x + 1)² = 4

Multiply out the left-hand side, move the 4 over, and solve by whatever method you prefer.

Eliz.