Find the approximate value of f(5.001) , where f(x) = x³ - 7x² + 15?

Take x as 5 and delta x as 0.001
Formula is f ( x + delta x) = f ( x ) + f'(x) * delta x
f'(x) = derivative of f ( x )
That's the method to find it. And don't do it by direct substitution.

f(5.001) = 5.001^3 - 7*5.001^2 + 15
= 125.075 - 175.070 + 15
= - 34.995

df = f'(x) dx
= (3x^2-14x) (.001) = (75-70)*(1/1000) = 5/1000
125-7*25+15+5/1000 = 125-175+15+5/1000=-35+5/1000 = -34.995

f ( x = 5.001) = x3 -7x2 + 15 ,
so f( 5.001 ) = ( 5.001)3 - 7 ( 5.001)2 + 15 = 125.075015- 7 x 25.010001+ 15 = - 34.994992 , that is around - 34.995 to the degree of three decimals

f(x) = x³ - 7x² + 15

f(5.001)
= f(5 + 0.001)
= (5 + 0.001)³ - 7(5 + 0.001)² + 15
= (5³ + 3×5²×0.001 + 3×5×0.001² + 0.001³) - 7(5² + 2×5×0.001 + 0.001²) + 15

The terms 3×5×0.001², 0.001³ and 0.001² are negligible. Hence,
f(5.001)
= (5³ + 3×5²×0.001) - 7(5² + 2×5×0.001) + 15
= (125 + 0.075) - 7(25 + 0.01) + 15
= 125 + 0.075 - 175 - 0.07 + 15
= -34.995

You have :
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f(x) = x^3 - 7x^2 + 15

f( x + delta x ) = f(x) + ( df/dx ) ( delta x )

f(5) = 5^3 - ( 7 ) ( 5^2 ) + 15

f(5) = 125 - 175 + 15 = - 35

df/dx = 3x^2 - 14x

df/dx = ( 3 ) ( 5 )^2 - ( 14 ) ( 5 ) = 75 - 70 = 5

( df/dx ) ( delta x ) = ( 5 ) ( 0.001 ) = 0.005

d(5.001)/dx = -35 + 0.005 = - 34.995 <--------

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