If f(x)=3x^{2}+15x+5,then the approximate val | vedclass

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(D) Let $x=3$ and $\Delta x=0.02$. Then,we have:$f(x+\Delta x) \approx f(x) + f'(x) \Delta x$Given $f(x) = 3x^{2} + 15x + 5$,we find the derivative:$f

Vedclass · Accepted Answer

$77$

Vedclass · Answer

(D) Let $x=3$ and $\Delta x=0.02$. Then,we have:$f(x+\Delta x) \approx f(x) + f'(x) \Delta x$Given $f(x) = 3x^{2} + 15x + 5$,we find the derivative:$f'(x) = \frac{d}{dx}(3x^{2} + 15x + 5) = 6x + 15$Now,substitute $x=3$ and $\Delta x=0.02$ into the approximation formula:$f(3.02) \approx f(3) + f'(3)(0.02)$Calculate $f(3)$:$f(3) = 3(3)^{2} + 15(3) + 5 = 3(9) + 45 + 5 = 27 + 45 + 5 = 77$Calculate $f'(3)$:$f'(3) = 6(3) + 15 = 18 + 15 = 33$Substitute these values back:$f(3.02) \approx 77 + 33(0.02)$$f(3.02) \approx 77 + 0.66$$f(3.02) \approx 77.66$Thus,the approximate value is $77.66$.

If $f(x)=3x^{2}+15x+5$,then the approximate value of $f(3.02)$ is (in $.66$)

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