For a given function y=f(x),delta y denotes t | vedclass

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(A) Given $y = f(x) = 2x^2 - 3x + 4$,$x = 5$,and $\delta x = 0.02$. First,calculate the actual error $\delta y = f(x + \delta x) - f(x)$: $\delta y =

Vedclass · Accepted Answer

$0.0008$

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(A) Given $y = f(x) = 2x^2 - 3x + 4$,$x = 5$,and $\delta x = 0.02$. First,calculate the actual error $\delta y = f(x + \delta x) - f(x)$: $\delta y = f(5.02) - f(5) = [2(5.02)^2 - 3(5.02) + 4] - [2(5)^2 - 3(5) + 4]$ $= [2(25.2004) - 15.06 + 4] - [50 - 15 + 4]$ $= [50.4008 - 15.06 + 4] - 39 = 39.3408 - 39 = 0.3408$. Next,calculate the differential $dy = f'(x) \cdot dx$: $f'(x) = \frac{d}{dx}(2x^2 - 3x + 4) = 4x - 3$. At $x = 5$,$f'(5) = 4(5) - 3 = 17$. $dy = 17 	imes 0.02 = 0.34$. Finally,the difference $\delta y - dy = 0.3408 - 0.34 = 0.0008$.

For a given function $y=f(x)$,$\delta y$ denotes the actual error in $y$ corresponding to an actual error $\delta x$ in $x$,and $dy$ denotes the approximate value of $\delta y$. If $y=f(x)=2x^2-3x+4$ and $\delta x=0.02$,then the value of $\delta y - dy$ when $x=5$ is

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