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Find the approximate value of f (5. 001)...

Find the approximate value of `f (5. 001)`, where `f(x)=x^3-7x^2+15`.

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`f(x) = x^(3) - 7x^(2) + 15`
` rArr f'(x) = 3x^(2) = 14 x`
Let `x = 5 and h= 0.001`
Now ` f(x+h) = f(x) + hf'(x)`
` rArr f(5.001) = f(x) + hf'(x)`
` = [5^(3) - 7 xx 5^(2) + 15]+0.001`
`xx[3 xx 5^(2) - 14 xx 5]`
` = - 35 + 0.001 xx 5`
` = - 35 + 0.005 =- 34.995`
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