WebNov 20, 2014 · For any y ∈ f ( f − 1 ( B), there exists x ∈ f − 1 ( B) satisfying y = f ( x). Clearly y = f ( x) ∈ B. For any y ∈ B, since f ( x) is an onto function, there is x ∈ X satisfying that f ( x) = y, i.e., x ∈ f − 1 ( y). So y = f ( x) ⊂ f ( f − 1 ( y)) ⊂ f ( f − 1 ( B)). Share Cite Follow edited Oct 23, 2024 at 15:04 rae306 9,402 3 18 46 Web(b) Find all x values not in the domain of y = (f f f)(x) Solution: (a) − 4 < 3 + 5 x < 4 − 7 < 5 x < 1 − 7 5 < 1 x < 1 5 − 5 7 > x > 5 5 < x < − 5 7 So the interval notation is x ∈ −∞, − 5 7 ∪ (5, ∞) (b) We have f (f (f (x))) = f f 3 + 5 x = f 3 + 5 3 + 5 x =3 + 5 3 + 5 3+ 5 x We have three denominators that cannot be ...
Find a real function $f:\\mathbb{R}\\to\\mathbb{R}$ such that $f(f(x ...
WebFree Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step WebJul 27, 2024 · For f 1 ( x), the only solution is f 1 ( x) = x. For f 2 ( x), the solutions are involutions. For f 3 ( x), the only answer is f 3 ( x) = x . For all other f n ( x), one solution is f n ( x) = x. My question: For n ≥ 3, is f n ( x) = x the only solution? If not, what are the solutions? Edit: @MattSamuel said that any involution works for an even n. jena thomas nitzsche
If \( f(x)=\frac{1}{1-x} \), show that, math …
WebConnecting f, f', and f'' graphically (video) Khan Academy Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere. WebFree math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Enter a problem... Algebra Examples Popular Problems Algebra Find f(f(x)) f(x)=3x-4 Setup the compositeresult function. Evaluateby substituting in the value of into . WebSep 2, 2014 · suppose f ( x) = x n can satisfy this equation. Then n − 1 = n 2. There are two complex solutions for this equation. I don't no they can be acceptable as an answers. But nothing seems like wrong. – Bumblebee Sep 2, 2014 at 12:35 @HagenvonEitzen The constant function f ( x) = 0 cannot have a 'root of order k '. jena university master application