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(A) Given function is $f(x) = x^2 - 2x + 1$. We need to find the approximate value at $x = 2.99$. Let $x = 3$ and $\Delta x = -0.01$,so that $x + \Del

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$3.96$

Vedclass · Answer

(A) Given function is $f(x) = x^2 - 2x + 1$. We need to find the approximate value at $x = 2.99$. Let $x = 3$ and $\Delta x = -0.01$,so that $x + \Delta x = 2.99$. The derivative is $f'(x) = 2x - 2$. Using the formula for linear approximation: $f(x + \Delta x) \approx f(x) + \Delta x \cdot f'(x)$. At $x = 3$,$f(3) = 3^2 - 2(3) + 1 = 9 - 6 + 1 = 4$. At $x = 3$,$f'(3) = 2(3) - 2 = 6 - 2 = 4$. Substituting these values into the approximation formula: $f(2.99) \approx f(3) + (-0.01) \cdot f'(3)$. $f(2.99) \approx 4 + (-0.01)(4)$. $f(2.99) \approx 4 - 0.04 = 3.96$.

Using differentiation,the approximate value of $f(x) = x^2 - 2x + 1$ at $x = 2.99$ is ....

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