Question

Show that the equation x³-5x-1=0 has a solution between x = 2 and x = 3. Find this solution correct to 1 decimal place. You must show your working.

pls help idk how to do this​

Answer 1

a)

f(x) = x³ - 5x - 1

f'(x) = (x³-5x-1)' = 2x² - 5

f'(x) = 0 <=> 3x² - 5 = 0

${x}^{2} = \dfrac{5}{3} \implies x = \pm \sqrt{\dfrac{5}{3}}$

if f'(x) < 0 then f(x) is decreasing

if f'(x) > 0 then f(x) is increasing

$x \in \Big(-\infty ; - \sqrt{\dfrac{5}{3}}) \to f(x) - increasing \\ x \in \Big(-\sqrt{\dfrac{5}{3}} ; \sqrt{\dfrac{5}{3}} \Big) \to f(x) - decreasing\\ x \in \Big(\sqrt{\dfrac{5}{3}} ; +\infty \Big) \to f(x) - increasing$

$\sqrt{\dfrac{5}{3}} < 2 < 3$

f(2) = 2³ - 10 - 1 = 8 - 11 = -3

f(3) = 3³ - 15 - 1 = 27 - 16 = 11

=> f(x) has a solution between x = 2 and x = 3

b)

using Newton-Raphson:

x₀=(2+3)/2=5/2=2.5

f(x₀)=f(2.5)=2.125

f'(x₀)=f'(2.5)=13.75

x₁=x₀-f(x₀)/f'(x₀)=2.3(45)

Approximate root of this equation using Newton Raphson method (1 decimal place) is 2.3