Webf ′ ( x) = 1 − 1 1 + x = x 1 + x, which says that f ( x) ≥ f ( 0) = 0. In another hand, let g ( x) = ln ( 1 + x) − x x + 1, where x > − 1. Thus, g ′ ( x) = 1 1 + x − 1 ( 1 + x) 2 = x ( 1 + x) 2, which gives again that x m i n = 0 and we obtain g ( x) ≥ g ( 0) = 0. Since, x ≠ 0 we get x x + 1 < ln ( 1 + x) < x. Done! Share WebUse Ne wton’s Meth od to det ermine 𝑥 1 and 𝑥 2 for the f unction 𝑓 (𝑥) = 𝑥 𝑐𝑜𝑠 (𝑥) − 𝑥 2 and the value of 𝑥 0 = 1 . Answer: 𝑥 1 = 0.8002329432 and 𝑥 2 = 0.744094398 5

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