Find approximate value of f (5.001) wher...

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

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Firstly break the number 5.001 as x=5 and △x=0.001 and use the relation f(x+△x)=f(x)+△x f′(x).
Consider f(x)=`x^3-7x^2+15`
`f'(x)=3x^2-14x`
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