Question

3. Use the mean value theorem to prove the following inequality. (1 +x)" >1 for z >0 andnEN

Answer 1

Mean value theorem: if f and f prime are continuous on [a, b] then f(b) - f(a)/b - a = f prime (c) For some value C between a and b. so its implies following three inequalities. If f prime(c) > 0, then f(b) > f(a) if f prime(c) < 0, then f(b) < f(a) if f prime (c) = 0, then f(b) = f(a) we need to prove the following inequality using MVT - (1 + x)^n > 1 + nx for x > 0 & n elementof N implies (1 + x)^n - (1 + nx) > 0 for x > 0 & n elementof N Let f(x) = (1 + x)^h - (1 + nx) we need to show that f(x) is always positive because f(0) = (1 + 0)^h - (1 + n middot 0) = 1 - 1 = 0 and f prime (x) = n(1 + x)^n - 1 - n f prime (x) =