If f x is a polynomial function satisfying
Web9 jan. 2015 · Let f be an integer polynomial with f ( x) f ( 1 / x) = f ( x) + f ( 1 / x). There are four solutions, f = 0, f = 2, and f = 1 ± x n. Clearly f = 0 is a solution. As in the question, … Web27 mrt. 2024 · If f(x) is a polynomial function satisfying f(x). f(1/x) = f(x) + f(1/x) ∀ x ∈ R – {0} and f(2) = 9, then find f (3). function; jee; jee mains; Share It On Facebook Twitter Email. 1 Answer +1 vote . answered Mar 27, 2024 …
If f x is a polynomial function satisfying
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Web7 mei 2015 · If f (x) is a polynomial satisfying 2 + f ( x) f ( y) = f ( x) + f ( y) + f ( x y), find f ( f ( 2)), given f ( 2) = 5. ATTEMPT:- f ( f ( 2)) = f ( 5), We can find f ( 0), f ( 1) and f ( 1 / 2) to be 1, 2 and 5 / 4 respectively. we can change the function to the form g ( x) ∗ g ( y) = g ( x y) by basic transformation. WebThe polynomial which satisfies f(x)f(1/x)=f(x)+f(1/x) is ±x n+1 (standard result) Given that f(2)=9 ⇒±2 n+1=9 ⇒2 n=8. (-ve sign not possible here) ⇒n=3. Hence the function is …
Web13 apr. 2024 · If \( f(x) \) is a polynomial function satisfying \( f(x) \cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right), \forall x \in R-\{0\} \) and \( f(2...
Web25 jun. 2024 · Suppose `f` is a real function satisfying `f(x+f(x))=4f(x)a n df(1)=4.` Then the value of `f(21)` is `16` `21` `64` `105` asked Jan 21, 2024 in Mathematics by EashtaBasu ( 97.0k points) class-12 Web21 mei 2024 · Find a Quadratic Function Satisfying Conditions on Derivatives Find a quadratic function f(x) = ax2 + bx + c such that f(1) = 3, f ′ (1) = 3, and f′′(1) = 2. Here, f ′ (x) and f′′(x) denote the first and second derivatives, respectively. Solution. Each condition required on f can be turned […]
Web6 mrt. 2024 · Given f (x) is a polynomial function of x, satisfying f (x) ⋅ f (y) = f (x) + f (y) + f (xy) − 2 and that f (2) = 5. Then f (3) is equal to? A) 10 B) 24 C) 15 D) none. Calculus …
WebDifferential equation f"(x) = f(x) f"(x) - f(x) = 0 [D 2-1] f(x) = 0. The corresponding auxiliary equation D 2 - 1 = 0. Root of auxiliary equation D 2 = 1. ⇒ D = \(±\) 1. Here, m = 1 and n = - 1. Roots are real and different. So complimentary solution is given by. ⇒ f(x) = c 1 e x + c 2 e-x. Put initial condition f(0) = 2 in the above ... tennessee technologyWebSolving polynomials We solve polynomials algebraically in order to determine the roots - where a curve cuts the \ (x\)-axis. A root of a polynomial function, \ (f (x)\), is a value... tennessee technological university costWebAnswer: The function satisfying the equation f(x)* f(1/x) = f(x) + f(1/x) takes the form f(x)=1+x^n and since it is given that f(4)=65, we obtain n=3 This turns the ... tennessee technology access program ttapWeb14 apr. 2024 · If \( f \) is a polynomial function satisfying \( f(x) \cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right) \) and \( f(2)=-3 \), then draw graph o... trey songz high schoolWebFirst assume P ( x) has positive degree. Therefore the limit as x goes to infinity cannot be 0 (as P ( x) is not bounded) so P ( x) must be constant. Finally, it is clear that the only constant polynomial satisfying the required condition is the 0 polynomial. Share Cite Follow answered Feb 23, 2013 at 22:43 fidbc 889 5 8 Add a comment tennessee technology center at livingstonWebIf f (x) is a polynomial function satisfying the condition f (x) × f (1x) = f (x) + f (1x) and f (2) = 9 then. Class 11. >> Applied Mathematics. >> Functions. >> Graphs of functions. … trey songz hometownWebFinal answer. Transcribed image text: Question: Let f (x) be a function satisfying f (0)= 0, f ′(0) = 4, f ′′(0) = −6 and f (3)(x) ≤ 12 for 0 ≤ x ≤ 1 Find the Taylor polynomial of degree 2 … trey songz i don\u0027t want to leave