Functions #13 - Questions and Answers

Question 1 How do you find domain restrictions? Answer 1 Identify the input values. Identify any restrictions on the input and exclude those values from the domain. Write the domain in interval form, if possible. Answer 2 The domain for some function f(x) consists of all the values of x you are allowed to plug in without "breaking" the rules of math. Question 2 Find the general solution of the given differential equation dy/dx - y/x = x²sin x give the largest interval over which the general solution is defined. Answer dy/dx - y/x = x²sin x dy/dx - (1/x)y = x²sin x This is 1-st order linear in y so integrating Factor = e^(∫-(1/x)dx) = e^(-lnx) = 1/x multiplying the given equation with I.F. we get (1/x)(dy/dx) - y/x² = x sin x d/dx (y/x) = x sin x Integrating both side we get y/x = ∫x sin x dx y/x = -x cos x + ∫cos x dx y/x = -x cos x + sin x + C y = -x²cos x + x sin x + Cx Since y is continuous on (-∞, ∞) so longest inter val of existence is (-∞,∞) Since xcos x and Cx both does not goes to 0 as x → ∞ Hence there is no transient form. Question 3 What is the inverse function of f(x) = x²? Answer 1 Hence f(x) = x² => y = x² => √y = √x² => √y = |x| => x = ±√y Answer 2 If we try to solve y = x² for x we do not get a single value. That means we do not get a function. We get x = ±√y In order to be invertible a function must be one-to-one. That means that we must have: for every x₁ ≠ x₂, we have f(-1) = f(1) (for example), so there is no inverse function. Question 4 Find the least integer n such that f(x) is O(xⁿ) for each of these functions. a) f(x) = 2x² + x³log x b) f(x) = 3x⁵ + (log x)⁴ c) f(x) = (x⁴ + x² + 1) / (x⁴ + 1) Answer 1 a) Given: f(x) = 2x² + x³log x When x > 0, we have the property log x ≤ x f(x) = 2x² + x³log x < 2x² + x⁴ The largest power of x mentioned in the definition of f(x) is the smallest n for which f(x) is O(xⁿ) n = 4 When x > 2, we have the property x² > x > 2 For convenience sake, we will choose k = 2 and thus use x > 2 |f(x)| = |2x² + x³log x| ≤ |2x²| + |x³log x| < |2x²| + |x⁴| = 2x² + x⁴ < x⁴ + x⁴ = 2x⁴ = 2|x⁴| Thus we need to choose C to be at least 2. Let us then take C = 2 By the definition of the Big-O notation, f(x) = O(x⁴) with k = 2 and C = 2 b) Given: f(x) = 3x⁵ + (log x)⁴ The largest power of x mentioned in the definition of f(x) is the smallest n for which f(x) is O(xⁿ) n = 5 When x > 0, we have the property log x ≤ x When x > 1, we also have the property x⁴ < x⁵ For convenience sake, we will choose k = 1 and thus us x > 1 |f(x)| = |3x⁵ + (log x)⁴| ≤ |3x⁵| + |(log x)⁴| = 3x⁵ + (log x)⁴ < 3x⁵ + x⁴ < 3x⁵ + x⁵ = 4x⁵ = 4|x⁵| Thus we need to choose C to be at least 4. Let us then take C = 4 By the definition of the Big-O notation f(x) = O(x⁵) with k = 1 and C = 4 c) Given: f(x) = (x⁴ + x² + 1) / (x⁴ + 1) Let us determine the quotient (with remainder) of the given using long division. The quotient is then 1 with remainder x²+ 1 and thus we can rewrite the given fractions as: f(x) = (x⁴ + x² + 1) / (x⁴ + 1) = 1 + (x²)/(x⁴+1) The largest power of x of the quotient is the smallest n for which f(x) is O(xⁿ). n = 0 For convenience sake, we will choose k = 0 and thus x > 0 |f(x) = 1 + x²/(x⁴ + 1) | ≤ |1| + |x²/(x⁴ + 1)| = 1 + (x²)/(x⁴) = 1 + 1/x² < 1 + 1 = 2 = 2*|1| Thus we need to choose C to be at least 2. Let us then take C=2 By the definition of the Big-O notation. f(x) = O(x⁰) = O(1) with k=0 and C=2 Answer 2 Answer: a) n = 3 b) n = 5 c) n = 4 Explanation: To find the least integer n such that f(x) is 0(xⁿ) for each of the given functions, we can analyze the highest degree terms in the functions. a) For f(x) = 2x² + x³log(x), the highest degree term is x³log(x). Since log(x) grows slower than any positive power of x, we can ignore it for our analysis. Therefore, the highest degree term is x³, and we can express f(x) as f(x) = 2x² + O(x³). Hence, n = 3. b) For f(x) = 3x⁵ + (log(x))⁴, the highest degree term is x⁵. The logarithmic term, (log(x))⁴, grows slower than any positive power of x. Thus, we can express f(x) as f(x) = 3x⁵ + O(x⁵). Therefore, n = 5. c) For f(x) = (x⁴ + x² + 1) / (x⁴ + 1), both the numerator and denominator have the highest degree of x⁴. When x tends to infinity, the other terms become negligible compared to the leading term. Thus, we can express f(x) as f(x) = x⁴/x⁴ + O(x⁰) = 1 + O(x⁰). Therefore, n = 4. Answer 3 a) For the function f(x) = 2x² + x³logx, we need to find the least integer n such that f(x) is O(xⁿ). Let's analyze the growth of each term separately: f(x) = 2x² + x³logx = O(x²) + O(x³logx) Since x³logx grows faster than x², we can ignore the O(x²) term. Therefore, we have: f(x) = O(x³logx) So, the least integer n such that f(x) is O(xⁿ) is n = 3. b) For the function f(x) = 3x⁵ + (logx)⁴, we need to find the least integer n such that f(x) is O(xⁿ). Similar to the previous example, let's analyze the growth of each term: f(x) = 3x⁵ + (logx)⁴ = O(x⁵) + O((logx)⁴) Since (logx)⁴ grows slower than x⁵, we can ignore the O((logx)⁴) term. Therefore, we have: f(x) = O(x⁵) Hence, the least integer n such that f(x) is O(xⁿ) is n = 5. c) For the function f(x) = (x⁴ + x² + 1) / (x⁴ + 1), we need to find the least integer n such that f(x) is O(xⁿ). Let's simplify the expression: f(x) = (x⁴ + x² + 1) / (x⁴ + 1) = (1 + 1/x² + 1/x⁴) / (1 + 1/x⁴) As x approaches infinity, the terms 1/x² and 1/x⁴ become negligible compared to 1. Thus, we have: f(x) = (1 + 0 + 0) / (1 + 0) = 1/1 = 1 Since f(x) is a constant value, it is O(x⁰) for any positive value of x. Therefore, the least integer n such that f(x) is O(xⁿ) is n = 0. To summarize: a) n = 3 b) n = 5 c) n = 0