Find the approximate value of f(2.01), where | vedclass

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(A) Let $x=2$ and $\Delta x=0.01.$ Then,we have:$f(x+\Delta x) \approx f(x) + f'(x) \Delta x$Given $f(x) = 4x^{2} + 5x + 2$,we find the derivative:$f'

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$28.21$

Vedclass · Answer

(A) Let $x=2$ and $\Delta x=0.01.$ Then,we have:$f(x+\Delta x) \approx f(x) + f'(x) \Delta x$Given $f(x) = 4x^{2} + 5x + 2$,we find the derivative:$f'(x) = 8x + 5$At $x=2$,$f(2) = 4(2)^{2} + 5(2) + 2 = 16 + 10 + 2 = 28$.At $x=2$,$f'(2) = 8(2) + 5 = 16 + 5 = 21$.Now,substitute these values into the approximation formula:$f(2.01) \approx f(2) + f'(2) \Delta x$$f(2.01) \approx 28 + (21)(0.01)$$f(2.01) \approx 28 + 0.21$$f(2.01) \approx 28.21$Thus,the approximate value of $f(2.01)$ is $28.21$.

Find the approximate value of $f(2.01),$ where $f(x)=4x^{2}+5x+2$.

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