The minimum value of the slope of the tangent | vedclass

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(B) The slope of the tangent to the curve $y=f(x)$ is given by $m = \frac{dy}{dx}$.Given $y = x^3 - 3x^2 + 2x + 93$,we differentiate with respect to $

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$-1$

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(B) The slope of the tangent to the curve $y=f(x)$ is given by $m = \frac{dy}{dx}$.Given $y = x^3 - 3x^2 + 2x + 93$,we differentiate with respect to $x$:$m = \frac{dy}{dx} = 3x^2 - 6x + 2$.To find the minimum value of the slope $m$,we differentiate $m$ with respect to $x$ and set it to zero:$\frac{dm}{dx} = 6x - 6$.Setting $\frac{dm}{dx} = 0$,we get $6x - 6 = 0$,which implies $x = 1$.Now,we check the second derivative to confirm the minimum:$\frac{d^2m}{dx^2} = 6 > 0$,which confirms that the slope has a minimum at $x = 1$.The minimum value of the slope is $m(1) = 3(1)^2 - 6(1) + 2 = 3 - 6 + 2 = -1$.

The minimum value of the slope of the tangent to the curve $y=x^3-3x^2+2x+93$ is

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