Approximate the real zeros of f(x)= x^3 - 4x + 6 to the nearest tenth.
Please select the best answer from the choices provided
A -5
B -2
C -5.2
D -2.5

x ≈ - 2.5; Option D

Step-by-step explanation:

Let us substitute f ( x ) with 0, as to find the " real zero " of the function;

0 = x^2 - 4x + 6, ⇒ Now find one solution for x^2 - 4x + 6 = 0, using Newton - Raphson, provided it applies an iterative process to approach one root of a function, ( x( n + 1 ) = x( n ) - ( f( x( n ) ) )/( f '( x( n ) ) ) ),

d / dx * ( x^3 - 4x + 6 ), ⇒ Apply sum / difference rule ( f ± g ) ' = f ' ± g, '

d / dx * ( x^3 ) - d / dx * ( 4x ) + d / dx * ( 6 ), ⇒ Simplify,

3x^2 - 4 + 0, 3x^2 - 4, ⇒ Compute x( n + 1 ) until Δ x( n + 1 ) < 0.000001,

x ≈ - 2.52510

Answer; x ≈ - 2.5; Option D

47.28 or 47.2857143

step-by-step explanation:

you just divide 331 by 7 and you get that answer. ( i think so tho)