Find the approximate value of f (5.001), where f (x) = x 3 − 7x 2 + 15.
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Let x = 5 and ∆x = 0.001
now f(x+∆x) = f(5.001)
∆y = f(x+∆x)-f(x)
=> f(x+∆x) = ∆y+f(x)
= f'(x).∆x + f(x) ......(°•°dy≈∆y and dx = ∆x)
= (3x² -14x).∆x + x³ - 7x² + 15
= (3(5)²-14(5))×0.001+(5³-7(5)²+15)
= (3×25-70)×0.001+(125-7×25+15)
= 5×0.001+(140-175)
= 0.005+(-35)
= -34.995
now f(x+∆x) = f(5.001)
∆y = f(x+∆x)-f(x)
=> f(x+∆x) = ∆y+f(x)
= f'(x).∆x + f(x) ......(°•°dy≈∆y and dx = ∆x)
= (3x² -14x).∆x + x³ - 7x² + 15
= (3(5)²-14(5))×0.001+(5³-7(5)²+15)
= (3×25-70)×0.001+(125-7×25+15)
= 5×0.001+(140-175)
= 0.005+(-35)
= -34.995
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