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

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(A) The given curve is $y = x^{3} - x + 1$. To find the slope of the tangent,we differentiate the equation with respect to $x$: $\frac{dy}{dx} = \frac

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(A) The given curve is $y = x^{3} - x + 1$. To find the slope of the tangent,we differentiate the equation with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^{3} - x + 1) = 3x^{2} - 1$. The slope of the tangent at a point $(x_{0}, y_{0})$ is given by the value of $\frac{dy}{dx}$ at that point. Given the $x$-coordinate $x_{0} = 2$,we substitute this into the derivative: $\left. \frac{dy}{dx} \right|_{x=2} = 3(2)^{2} - 1$. Calculating the value: $3(4) - 1 = 12 - 1 = 11$. Thus,the slope of the tangent at $x = 2$ is $11$.

Find the slope of the tangent to the curve $y = x^{3} - x + 1$ at the point where the $x$-coordinate is $2$.

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