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If f(x) = 3x ^2 + 15x + 5, then the approximate value of f (3.02) is

8) If f(x) = 3x ^2 + 15x + 5, then the approximate value of f (3.02) is
(A) 47.66                   (B) 57.66                       (C) 67.66                         (D) 77.66

Answers (1)
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Let x = 3 and \Delta x = 0.02
f(x+\Delta x) = 3(x+\Delta x)^2 +15(x+\Delta x)+5
\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 = (6x+15).(0.02) \ \ \ \ \ \ \ \ \ (\because y = f(x) = 3x^2+15x+5 \ and \ \Delta x = 0.02)\\ dy = 0.12x+0.3
f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.12x + 0.3 + 3x^2 + 15x +5\\ f(x+\Delta x) = 0.12(3)+0.3+3(3)^2+15(3)+5\\ f(x+\Delta x) = 0.36+ 0.3 + 27 + 45 + 5\\ f(x+\Delta x) = 77.66
Hence,  the approximate value of f (3.02) is 77.66
Hence, (D) is the correct answer

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