The approximate value of y=(1.01)^3+2(1.01)^{ | vedclass

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(A) Let the function be $f(x) = x^3 + 2x^{3/2} + 5$. We need to find the approximate value at $x = 1.01$. Let $x = 1$ and $\Delta x = 0.01$. The diffe

Vedclass · Accepted Answer

$8.06$

Vedclass · Answer

(A) Let the function be $f(x) = x^3 + 2x^{3/2} + 5$. We need to find the approximate value at $x = 1.01$. Let $x = 1$ and $\Delta x = 0.01$. The differential of $y$ is given by $dy = f'(x) \Delta x$. First,find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(x^3 + 2x^{3/2} + 5) = 3x^2 + 2 \cdot \frac{3}{2} x^{1/2} = 3x^2 + 3x^{1/2}$. At $x = 1$,$f'(1) = 3(1)^2 + 3(1)^{1/2} = 3 + 3 = 6$. The value of the function at $x = 1$ is $f(1) = 1^3 + 2(1)^{3/2} + 5 = 1 + 2 + 5 = 8$. Using the approximation formula $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$: $f(1.01) \approx f(1) + f'(1) \Delta x = 8 + 6(0.01) = 8 + 0.06 = 8.06$.

The approximate value of $y=(1.01)^3+2(1.01)^{\frac{3}{2}}+5$ is

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