Find the slope of the tangent to the curve y | vedclass

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(A) The given curve is $y = x^{3} - 3x + 2$. To find the slope of the tangent,we differentiate the equation with respect to $x$: $\frac{dy}{dx} = \fra

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(A) The given curve is $y = x^{3} - 3x + 2$. To find the slope of the tangent,we differentiate the equation with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^{3} - 3x + 2) = 3x^{2} - 3$. The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx}$. We need to find the slope at the point where the $x$-coordinate is $3$. Substituting $x = 3$ into the derivative: $\left. \frac{dy}{dx} \right|_{x=3} = 3(3)^{2} - 3 = 3(9) - 3 = 27 - 3 = 24$. Thus,the slope of the tangent at $x = 3$ is $24$.

Find the slope of the tangent to the curve $y = x^{3} - 3x + 2$ at the point whose $x$-coordinate is $3$.

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