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

Explanation

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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