If frac{dy}{dx} = (x - 1)^3 (x - 2)^4,then y | vedclass

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(C) Given $\frac{dy}{dx} = (x - 1)^3 (x - 2)^4$.To find critical points,set $\frac{dy}{dx} = 0$,which gives $x = 1$ and $x = 2$.We use the first deriv

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a minimum at $x = 1$

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(C) Given $\frac{dy}{dx} = (x - 1)^3 (x - 2)^4$.To find critical points,set $\frac{dy}{dx} = 0$,which gives $x = 1$ and $x = 2$.We use the first derivative test to check the nature of these points.For $x = 1$:- If $x = 1 - h$ (where $h > 0$ is very small),$\frac{dy}{dx} = (1 - h - 1)^3 (1 - h - 2)^4 = (-h)^3 (-1 - h)^4 = (-h^3)(+) = -ve$.- If $x = 1 + h$,$\frac{dy}{dx} = (1 + h - 1)^3 (1 + h - 2)^4 = (h)^3 (h - 1)^4 = (+)(+) = +ve$.Since $\frac{dy}{dx}$ changes sign from negative to positive at $x = 1$,$y$ has a local minimum at $x = 1$.For $x = 2$:- If $x = 2 - h$,$\frac{dy}{dx} = (2 - h - 1)^3 (2 - h - 2)^4 = (1 - h)^3 (-h)^4 = (+)(+) = +ve$.- If $x = 2 + h$,$\frac{dy}{dx} = (2 + h - 1)^3 (2 + h - 2)^4 = (1 + h)^3 (h)^4 = (+)(+) = +ve$.Since $\frac{dy}{dx}$ does not change sign at $x = 2$,$x = 2$ is a point of inflection.

If $\frac{dy}{dx} = (x - 1)^3 (x - 2)^4$,then $y$ has:

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