Midterm 2 with Solution - Elementary Real Analysis | MATH 444, Exams of Mathematical Methods for Numerical Analysis and Optimization

Download Midterm 2 with Solution - Elementary Real Analysis | MATH 444 and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! MATH 444: SOLUTIONS FOR MIDTERM 2 1 (10 points): Compute lim x→1 √ x2 + 3 − 2 x − 1 Note that, for x 6= 1, √ x2 + 3 − 2 x − 1 = √ x2 + 3 − 2 x − 1 · √ x2 + 3 + 2√ x2 + 3 + 2 = x2 − 1 (x − 1)( √ x2 + 3 + 2) = (x − 1)(x + 1) (x − 1)( √ x2 + 3 + 2) = x + 1√ x2 + 3 + 2 , hence lim x→1 √ x2 + 3 − 2 x − 1 = limx→1 x + 1√ x2 + 3 + 2 = 1 + 1√ 12 + 3 + 2 = 1 2 . 2 (10 points): For x ∈ R, define f(x) = x3 + 4x + 1. (a) Prove that f is strictly increasing on R. f ′(x) = 3x2 + 4 > 0 for any x ∈ R. Thus, f is strictly increasing. (b) Consider g = f−1 (the inverse function to f). Compute g′(6). Hint. g(6) is an integer between −2 and 2. By trial and error, we see that f(1) = 6, hence g(6) = 1. Then g′(6) = 1 f ′(g(6)) = 1 f ′(1) = 1 7 . 3 (10 points): For x ∈ R, define f(x) = x cos x. Prove that the function f is not uniformly continuous on R. Suppose, for the sake of contradiction, that f is uniformly continuous. Then there exists δ > 0 s.t. |f(x) − f(y)| < 1 whenever |x − y| < δ. Pick N ∈ N s.t. π/N < δ. Fix M > N . For 0 ≤ k ≤ N , let xk = 2πM + kπ/N . Then x0 = 2πM , and xN = 2πM +π, hence f(xN )−f(x0) = (4M+1)π. On the other hand, |xk−xk−1| < δ for 1 ≤ k ≤ N , hence |f(xk) − f(xk−1)| < 1, and, by the triangle inequality, |f(xN) − f(x0)| ≤ N∑ k=1 |f(xk) − f(xk−1)| < N < M, a contradiction. 1