Find the approximate value of f(5.001), where | vedclass

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(A) Let $x=5$ and $\Delta x=0.001$. Then,we have:$f(5.001) = f(x+\Delta x) \approx f(x) + f'(x) \Delta x$.First,calculate $f(x)$ at $x=5$:$f(5) = (5)^

Vedclass · Accepted Answer

$-34.995$

Vedclass · Answer

(A) Let $x=5$ and $\Delta x=0.001$. Then,we have:$f(5.001) = f(x+\Delta x) \approx f(x) + f'(x) \Delta x$.First,calculate $f(x)$ at $x=5$:$f(5) = (5)^{3} - 7(5)^{2} + 15 = 125 - 175 + 15 = -35$.Next,find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^{3} - 7x^{2} + 15) = 3x^{2} - 14x$.Calculate $f'(x)$ at $x=5$:$f'(5) = 3(5)^{2} - 14(5) = 3(25) - 70 = 75 - 70 = 5$.Now,use the approximation formula:$f(5.001) \approx f(5) + f'(5) \Delta x$$f(5.001) \approx -35 + (5)(0.001)$$f(5.001) \approx -35 + 0.005 = -34.995$.Hence,the approximate value of $f(5.001)$ is $-34.995$.

Find the approximate value of $f(5.001),$ where $f(x)=x^{3}-7x^{2}+15$.

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