Define an anonymous function that can compute f(x) = x³ – x – 1. Underneath the function definition, call the function to compute y = f(x) for 0 < x< 4
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- Which functions are one-to-one? Which functions are onto? Describe the inversefunction for any bijective function.(a) f : Z → N where f is defined by f (x) = x4 + 1(b) f : N → N where f is defined by f (x) = { x/2 if x is even, x + 1 if x is odd}(c) f : N → N where f is defined by f (x) = { x + 1 if x is even, x − 1 if x is odd}Simplify the complement of the following function: F(A,B,C,D)=(0,2,4,5,8,9,10,11) Your answer: F=((A'B'D)' (BC)'(AB)')' F=((A'BD)'(BC)'(AB)')' F=((A'B'D)'(B'C)'(AB)') F=((A'B'D')' (BC)'(AB)')For each of the following pairs of functions f(n) and g(n), determine whether f(n) = O(g(n)), g(n) = O(f(n)), or not. Justify.(a) ?(?)=( ?2−?) / 2 ;?(?)=6?(b) ?(?)=?∗log(?); ?(?)=?√? / 2
- We have been working extensively with the predicate "eventually greater than" defined on pairs of functions f of g. Which of the following is equivalent to f(x) is not eventually greater than g(x)? (Select all that apply) Group of answer choices ¬((∃x0)(∀x)(x>x0→f(x)>g(x))) ((∃x0)(∀x)(x>x0→f(x)≤g(x))) (∀x)(∃x0)(x0>x→f(x0)≤g(x0)) (∀x)(∃x0)(x0>x→f(x0)≤g(x0))Consider the function f : N × N → N given byf(m, n) = 2m-1(2n − 1), (m, n) ∈ N × NShow that f is bijectiveThe sequence of values: n, c(n), c(c(n)), c(c(c(n))), c(c(c(c(n)))),... is known as the hailstone sequence starting at n. Implement c(n) as a function. Use predicates to test the parity of each input n.
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- % Define the function f(x) f = @(x) -1*(x < 0) + 1*(x >= 0 & x <= 2); % Set the maximum value of N Nmax = 512; % Initialize the x-axis values for plotting x = linspace(-1,3,1000); % Compute the Fejér sums for various values of N F = zeros(length(x), Nmax); for N = 1:Nmax % Compute the Nth Fejér sum Sn = @(x) 0; for n = 1:N bn = (-1)^(n+1) / (n*pi); Sn = @(x) Sn(x) + bn*sin(n*pi*x); end FN = @(x) (1/N) * Sn(x); % Evaluate the Fejér sum at each point on the x-axis I tried to submit this to MATLAB, but can't seem to get it right.The question describes a function S(k) which is defined as the sum of the positive divisors of a positive integer k, minus k itself. The function S(1) is defined as 1, and for any positive integer k greater than 1, S(k) is calculated as S(k) = σ(k) - k, where σ(k) is the sum of all positive divisors of k. Some examples of S(k) are given: S(1) = 1 S(2) = 1 S(3) = 1 S(4) = 3 S(5) = 1 S(6) = 6 S(7) = 1 S(8) = 7 S(9) = 4 The question then introduces a recursive sequence a_n with the following rules: a_1 = 12 For n ≥ 2, a_n = S(a_(n-1)) Part (a) of the question asks to calculate the values of a_2, a_3, a_4, a_5, a_6, a_7, and a_8 for the sequence. Part (b) modifies the sequence to start with a_1 = k, where k is any positive integer, and the same recursion formula applies: for n ≥ 2, a_n = S(a_(n-1)). The question notes that for many choices of k, the sequence a_n will eventually reach and remain at 1, but this is not always the case. It asks to find, with an explanation, two specific…Write a function that, given a list of non-overlapping intervals of items, takes an item as an input and identifies which, if any, interval that item lies in. If the items are integers and the intervals are 1643-2033, 5532-7643, 8999-10332, 5666653-5669321, then the query point 9122 is in the third interval while 8122 is not.