The local maximum of y=x^3-3 x^2+5 is attaine | vedclass

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(A) Given, $y=x^3-3 x^2+5$. On differentiating both sides with respect to $x$, we get $\frac{dy}{dx} = 3x^2 - 6x$. For local maxima or local minima, w

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$x=0$

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(A) Given, $y=x^3-3 x^2+5$. On differentiating both sides with respect to $x$, we get $\frac{dy}{dx} = 3x^2 - 6x$. For local maxima or local minima, we set $\frac{dy}{dx} = 0$. $3x^2 - 6x = 0 \Rightarrow 3x(x-2) = 0 \Rightarrow x = 0$ or $x = 2$. Now, differentiating $\frac{dy}{dx}$ with respect to $x$, we get $\frac{d^2y}{dx^2} = 6x - 6$. At $x = 0$, $\frac{d^2y}{dx^2} = 6(0) - 6 = -6 Since the second derivative is negative at $x = 0$, $x = 0$ is a point of local maxima. At $x = 2$, $\frac{d^2y}{dx^2} = 6(2) - 6 = 6 > 0$. Since the second derivative is positive at $x = 2$, $x = 2$ is a point of local minima. Therefore, the local maximum is attained at $x = 0$.

The local maximum of $y=x^3-3 x^2+5$ is attained at

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