If y=2x^3-2x^2+3x-5, then for x=2 and Delta x | vedclass

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(B) Given the function $y = 2x^3 - 2x^2 + 3x - 5$. To find the approximate change $\Delta y$, we use the formula $\Delta y \approx \frac{dy}{dx} \cdot

Vedclass · Accepted Answer

$1.9$

Vedclass · Answer

(B) Given the function $y = 2x^3 - 2x^2 + 3x - 5$. To find the approximate change $\Delta y$, we use the formula $\Delta y \approx \frac{dy}{dx} \cdot \Delta x$. First, differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(2x^3 - 2x^2 + 3x - 5) = 6x^2 - 4x + 3$. Now, evaluate the derivative at $x = 2$: $\left(\frac{dy}{dx}ight)_{x=2} = 6(2)^2 - 4(2) + 3 = 6(4) - 8 + 3 = 24 - 8 + 3 = 19$. Given $\Delta x = 0.1$, we calculate $\Delta y$: $\Delta y \approx 19 	imes 0.1 = 1.9$.

If $y=2x^3-2x^2+3x-5$, then for $x=2$ and $\Delta x=0.1$, the value of $\Delta y$ is:

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