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

Explanation

Let `f(x)= x^3-7x^2+15`

and `x=5`

`x+trianglex=5.001`

`impliestrianglex=0.001`

For `x=5,`

`y=125-175+15=-45`

Let `dx=trianglex=0.001`

As`y=x^3-7x^2+15`,

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