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Analysis Homework #2 Solutions 1. Show that the set A = {x + 1 x : x > 0} is such that inf A = min A = 2. • Assuming that x > 0, one easily finds that x+ 1 ≥2 x ⇐⇒ x2 + 1 ≥ 2x ⇐⇒ (x − 1)2 ≥ 0. Now, the rightmost inequality is obviously true and equality holds when x = 1, so the leftmost inequality is also true and equality holds when x = 1. This shows that 2 is the least element of A, namely that min A = 2. Once a minimum is known to exist, the infimum also does and the two are equal; thus, inf A = 2 as well. 2. Show that the set B = {x ∈ R : |2x − 3| < 1} has no maximum. • First of all, note that the given inequality is equivalent to |2x − 3| < 1 ⇐⇒ ⇐⇒ ⇐⇒ −1 < 2x − 3 < 1 2 < 2x < 4 1 < x < 2. If B has a maximum, then 1 < max B < 2 and the average y = max B+2 2 satisfies 1 < max B < y < 2. This makes y an element of B which is bigger than max B, a contradiction. 3. Let f be the function defined by the formula x3 − 8x − 3 f (x) = x−3 for all x ̸= 3. Determine the minimum value attained by f . • Using division of polynomials, we see that f (x) is given by f (x) = x3 − 8x − 3 = x2 + 3x + 1. x−3 Once we now complete the square, we easily find that ( )2 ( )2 ( )2 3 3 3 5 2 · 3x 2 + − +1= x− − . f (x) = x + 2 2 2 2 4 Since squares are always non-negative, this actually implies that min f (x) = −5/4. 4. Use the ε-δ definition of limits to show that lim (4x − 3) = 5. x→2 • Let ε > 0 be given and note that |4x − 3 − 5| = |4x − 8| = 4|x − 2|. Taking δ = ε/4, we conclude that |x − 2| < δ =⇒ |4x − 3 − 5| = 4|x − 2| < 4δ = ε.