Find the approximate value of f(3.02), where | vedclass

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Question

(A) Let $x=3$ and $\Delta x=0.02$. Then,the approximate value is given by $f(x+\Delta x) \approx f(x) + f'(x) \Delta x$.First,calculate $f(x)$ at $x=3

Vedclass · Accepted Answer

$45.46$

Vedclass · Answer

(A) Let $x=3$ and $\Delta x=0.02$. Then,the approximate value is given by $f(x+\Delta x) \approx f(x) + f'(x) \Delta x$.First,calculate $f(x)$ at $x=3$:$f(3) = 3(3)^{2} + 5(3) + 3 = 3(9) + 15 + 3 = 27 + 15 + 3 = 45$.Next,find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(3x^{2} + 5x + 3) = 6x + 5$.Now,calculate $f'(x)$ at $x=3$:$f'(3) = 6(3) + 5 = 18 + 5 = 23$.Using the approximation formula:$f(3.02) \approx f(3) + f'(3) \Delta x$$f(3.02) \approx 45 + (23)(0.02)$$f(3.02) \approx 45 + 0.46 = 45.46$.Thus,the approximate value of $f(3.02)$ is $45.46$.

Find the approximate value of $f(3.02),$ where $f(x)=3x^{2}+5x+3$.

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