Question

Find the approximate value of f (5.001), where f ( x ) = x 3 − 7 x 2 + 15.

Answer 1

The given function f( x ) is defined as,

f( x )= x 3 −7 x 2 +15

For the given function f( x+Δx ),

f( x+Δx )= ( x+Δx ) 3 −7 ( x+Δx ) 2 +15

Let, Δy be the approximate error which is defined as,

Δy=f( x+Δx )−f( x ) f( x+Δx )=f( x )+Δy (1)

For a function f( x ) we know that,

Δy Δx = f ′ ( x ) Δy= f ′ ( x )×Δx

Substitute the values in equation (1).

f( x+Δx )=f( x )+ f ′ ( x )×Δx(2)

Differentiate the given function with respect to x.

f ′ ( x )= d( x 3 −7 x 2 +15 ) dx =3 x 2 −14x

Use equation (2) for f( x )= x 3 −7 x 2 +15.

f ′ ( x+Δx )=( x 3 −7 x 2 +15 )+( 3 x 2 −14x )⋅Δx(3)

Substitute 5 for x and 0.001 for Δx in equation (3).

f( 5+0.001 )=[ 5 3 −7 ( 5 ) 2 +15 ]+[ 3 ( 5 ) 2 −14( 5 ) ]( 0.001 ) =[ 125−175+15 ]+( 75−70 )( 0.001 ) =−35+5( 0.001 ) =−34.995

Thus, the approximate value of f( 5.001 ) is −34.995.