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f(x) = ex - 2x - 1 = 0 function's [1; 2] there's a root within...

f(x) = ex - 2x - 1 = 0 function's [1; 2] there's a root within the closed range Show. Graphs f1 (x) = ex and f2 (x) = 2x + 1 functions on an equivalent axis Indicate the situation of the basis by drawing. Functional iteration by taking x0 = 1.5 (sequential approaches) method with 5 decimal precision (5D).

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