Question #151975
Trace the curve x^3 − 6x^2 + 11x − y = 6.
1
Expert's answer
2020-12-21T19:33:46-0500

Equation of the curve is y=x36x2+11x6y = x^3-6x^2+11x-6


This equation can be written as,

y=(x1)(x2)(x3)y = (x-1)(x-2)(x-3)

When x=0, y=-6.


So, I can directly find the x where y=0,

It will give me, x=1,2,3x= 1,2,3

So I will have intervals as (,1],[1,2],[2,3],[3,)(-\infty, 1], [1,2],[2,3],[3,\infty)


In interval (,1](-\infty, 1], If I put any value from -\infty to 1, value of the y will increase.

In interval [3,)[3, \infty) , on increasing value of x, y will increase.


Now, dydx=3x212x+11\frac{dy}{dx} = 3x^2-12x+11

Putting dydx=0,\frac{dy}{dx} =0, We get, x=1.423,2.577x = 1.423,2.577


d2ydx2=6x12\frac{d^2 y}{dx^2} = 6x-12


At x=1.423, d2ydx2=6(1.423)12<0\frac{d^2 y}{dx^2} = 6(1.423)-12 < 0, Hence it is point of maxima.


At x=2.577, d2ydx2=6x12>0\frac{d^2 y}{dx^2} = 6x-12 > 0 , So it is point of minimum.


Using all these properties, We find that curve will look like,





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