Answer
a) Intermediate Value Theorem: since f is a polynomial and continuous, and f ( 2 ) < 0 and f ( 3 ) > 0, then there exists a number such that f ( c ) = 0 .
b) $x_{5}$ ≈ $x_{6}$ ≈ 2.630020
Work Step by Step
Work for a:
f ( 2 ) = 3$( 2 )^{4}$ – 8( 2 )³ + 2
f ( 2 ) = 3*16 – 8*8 + 2
f ( 2 ) = 48 – 64 + 2
f ( 2 ) = - 14
f ( 3 ) = 3$( 3 )^{4}$ – 8( 3 )³ + 2
f ( 3 ) = 3 * 81 – 8 * 27 + 2
f ( 3 ) = 243 – 216 + 2
f ( 3 ) = 29
a) Intermediate Value Theorem: since f is a polynomial and continuous, and f ( 2 ) < 0 and f ( 3 ) > 0, then there exists a number such that f ( c ) = 0 .
Work for b:
f ( x ) = 3$x^{4}$ – 8x³ + 2
f ‘( x ) = 12x³ – 24x²
f ‘( x ) = 12x²( x – 2 )
$x_{2}$ = $x_{1}$ - $\frac{3x^{4} – 8x³ + 2}{12x²( x – 2 )}$
Since the root is between f ( 2 ) and f ( 3 ) , let $x_{1}$ = 2.5
$x_{2}$ = ( 2.5 ) - $\frac{3( 2.5 )^{4} – 8( 2.5 )³ + 2}{12( 2.5 )²( ( 2.5 ) – 2 )}$
$x_{2}$ = 2.5 - $\frac{-5.8125}{37.5}$
$x_{2}$ = 2.5 - ( - 0.155 )
$x_{2}$ = 2.655
$x_{3}$ = $x_{2}$ - $\frac{3x^{4} – 8x³ + 2}{12x²( x – 2 )}$
$x_{3}$ = ( 2.655 ) - $\frac{3( 2.655 )^{4} – 8( 2.655 )³ + 2}{12( 2.655 )²( ( 2.655 ) – 2 )}$
$x_{3}$ = ( 2.655 ) - $\frac{1.344969}{55.405337}$
$x_{3}$ = 2.630725
$x_{4}$ = $x_{3}$ - $\frac{3x^{4} – 8x³ + 2}{12x²( x – 2 )}$
$x_{4}$ = ( 2.630725 ) - $\frac{3( 2.630725 )^{4} – 8( 2.630725 )³ + 2}{12( 2.630725 )²( ( 2.630725 ) – 2 )}$
$x_{4}$ = 2.630725 – 0.000704
$x_{4}$ = 2.630021
$x_{5}$ = $x_{4}$ - $\frac{3x^{4} – 8x³ + 2}{12x²( x – 2 )}$
$x_{5}$ = ( 2.630021 ) - $\frac{3( 2.630021 )^{4} – 8( 2.630021 )³ + 2}{12( 2.630021 )²( ( 2.630021 ) – 2 )}$
$x_{5}$ = ( 2.630021 ) - $\frac{0.000039}{52.294342}$
$x_{5}$ = 2.630021 – 0.000001
$x_{5}$ = 2.630020
$x_{6}$ = $x_{5}$ - $\frac{3x^{4} – 8x³ + 2}{12x²( x – 2 )}$
$x_{6}$ = ( 2.630020 ) - $\frac{3( 2.630020 )^{4} – 8( 2.630020 )³ + 2}{12( 2.630020 )²( ( 2.630020 ) – 2 )}$
$x_{6}$ = ( 2.630020 ) - $\frac{-0.000013}{52.29422}$
$x_{6}$ = ( 2.630020 ) – ( - 0.000000 )
$x_{6}$ = 2.630020
b) $x_{5}$ ≈ $x_{6}$ ≈ 2.630020
