Given f(x) = x^3 - 4x,if x changes from 2 to | vedclass

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(B) Let $y = f(x) = x^3 - 4x$. We are given $x = 2$ and the change in $x$,denoted by $\Delta x = 1.99 - 2 = -0.01$. The approximate change in $y$,deno

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$-0.08$

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(B) Let $y = f(x) = x^3 - 4x$. We are given $x = 2$ and the change in $x$,denoted by $\Delta x = 1.99 - 2 = -0.01$. The approximate change in $y$,denoted by $\Delta y$,is given by $\Delta y \approx \frac{dy}{dx} \Delta x$. First,find the derivative of $f(x)$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^3 - 4x) = 3x^2 - 4$. Now,evaluate the derivative at $x = 2$: $\left. \frac{dy}{dx} ight|_{x=2} = 3(2)^2 - 4 = 3(4) - 4 = 12 - 4 = 8$. Finally,calculate the approximate change $\Delta y$: $\Delta y \approx 8 	imes (-0.01) = -0.08$.

Given $f(x) = x^3 - 4x$,if $x$ changes from $2$ to $1.99$,then the approximate change in the value of $f(x)$ is

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