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

Answers (1)

Let x = 5 and \Delta x = 0.001
f(x+\Delta x) =(x+\Delta x)^3 - 7(x+\Delta x)^2 +15
\Delta y = f(x+\Delta x) - f(x)\\ f(x+\Delta x) = \Delta y + f(x)
We know that \Delta y is approximately equal to dy
dy = \frac{dy}{dx}.\Delta x\\ dy = (3x^2 - 14x).(0.001) \ \ \ \ \ \ \ \ \ (\because y = f(x) = x^3-7x^2+15 \ and \ \Delta x = 0.001)\\ dy =0.003x^2 -0.014x
f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.003x^2 - 0.014x + x^3 - 7x^2 +15\\ f(x+\Delta x) =0.003(5)^2-0.014(5)+(5)^3-7(5)^2+15\\ f(x+\Delta x) = 0.075-0.07+125-175+15\\ f(x+\Delta x) = -34.995
Hence, the approximate value of f (5.001), where f (x) = x^3 - 7x^2 + 15\ is \ -34.995

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