Statement II is not a correctUse binomial theorem ( a − b) 2 = a 2 − 2 a b b 2 to expand ( x − 2) 2 fx=9\left (x^ {2}4x4\right) ∣ f ∣ x = 9 − ( x 2 − 4 x 4) To find the opposite of x^ {2}4x4, find the opposite of each term To find the opposite of x 2 − 4 x 4, find the opposite of each term fx=9x^ {2}4x4Now suppose that x1,x2 ∈ Z and f(x1) = f(x2) Then x1 5 = x2 5 and thus x1 = x2 It follows that f is onetoone Consequently, f is a bijection Notice that if a function f X → Y is a onetoone correspondence, then it associates one and only one value of y to each value in x In particular, it makes sense to define a reverse function
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Let f(x 1/x)=x^2 1/x^2 then f(x) is-It has been provided or mentioned that a function F(x1)=x^2–3x2 Interestingly, if we substitute a value of x1 in the place of x, we obtain, F((x1)1)=F(x) Therefore, F(x)=(x1)^2–3(x1)2=x^2–2x1–3x32=x^2–5x6 Hence, the function F(x) canM 2, and M 3 be metric spaces Let gbe a uniformly continuous function from M 1 into M 2, and let fbe a uniformly
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2) If fn → f uniformly on E, then fn → f pointwise on E The sequence f n ( x )= x n on 0 , 1 discussed in Example 1 of the previous section shows thatStatement II is a correct explanation of Statement I (c) Statement I is true, Statement II is true;1 1 and a = b = 2, then we see that a (b ~u) = 2 2 1 1 = 2 4 4 = 16 16 while (ab) ~u = 4 1 1 = 8 8 so we see that a (b ~u) 6= ( ab) ~u (c) V is the set of functions from R to the positive real numbers that is, V is the set of functions f with domain R such that f(x) > 0 for all x 2R Let f and g be in V We de ne f g and c f by de ning
(b) Pick = 1 Given any >0, pick x>0 such that 3 x2 2 >1 Then d(x 2;x) < but we have d(f(x 2);f(x)) = j(x 2)3 x3j= j 3 x2 2 3 2x 22 3 23 j 3 x2 2 >1 This shows that f(x) = x3 is not uniformly continuous on R 445 Let M 1;If f (x) = x1/x1 then find the value of f (2x) Find the answer to this question along with unlimited Maths questions and prepare better for JEE examination31 Continuity 23 so given ϵ > 0, we can choose δ = √ cϵ > 0 in the definition of continuity To prove that f is continuous at 0, we note that if 0 ≤ x < δ where δ = ϵ2 > 0, then f(x)−f(0) = √ x < ϵ Example 38 The function sin R → R is continuous on R To prove this, we use the trigonometric identity for the difference of sines and the inequality sinx ≤ x
Distinct solutions c 1;c 2 and c 3 and nally once more to conclude that f000(x) = 0 has at least two solutions d 1 and d 2 Next, note that since f is a quartic polynomial, f0must be a cubic, f00must be a quadratic and f000must be a linear polynomialWe therefore have a linearX0= 2 1 1 3 x et t 1 This becomes the equations x0 1= 2x x 2 tet x0 2= x 1 3x et 9416 Determine whether the given vector functions are linearly dependent or independent on the interval (1 ;1) sint cost ; Explanation By definition of the derivative f '(x) = lim h→0 f (x h) − f (x) h So with f (x) = 1 x2 we have;
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Let f (x) = (x 1)2 1, x ≥ 1 Statement I The set {x f (x) = f1(x)} = {0, 1} Statement II f is a bijection (a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is true;Let f x = x x then f '0 is equal to1 0 1 2 Please scroll down to see the correct answer and solution guideOf course, the simple answer would be to just write ( x − 2) for every x you see The answer would be f ( x − 2) = ( x − 2) 2 − 3 ( x − 2) 1 f ( x − 2) = x 2 − 4 x 4 − ( 3 x − 6) 1 f ( x − 2) = x 2 − 4 x 4 − 3 x 6 1 f ( x − 2) = x 2 − 7 x 11
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Sin2t cos2t We compute the Wronskian det sint sin2t cost cos2t = sintcos2t sin2tcost= sint where the last step can be deduced byAnswer to Let f(x) = (x^2 3x 5)(sqrt(x) 1/fifth root of x) Find f'(x) By signing up, you'll get thousands of stepbystep solutions toGiven a function g with this property, we can easily construct a suitable f Just let f ( x) = { g ( x) x ≥ 0 g ( − x) x < 0 If g is additionally continuous then so is f We can find a lot of continuous g Pick a 1 ∈ ( 0, 1), let a 0 = 0 and recursively a n = a n − 2 2 1 for n ≥ 2
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Get an answer for '`f(x) = x/(x^2 1)` (a) Find the intervals on which `f` is increasing or decreasing (b) Find the local maximum and minimum values of `fLet p ∈ Z be any prime We will show that hx2 1i is properly contained in hx2 1,pi which is not equal to Zx This will prove that hx2 1i is not maximal Since every nonzero element of hx2 1i has degree at least 2, p 6∈ hx2 1i This proves that hx2 1i is properly contained in hx2 1,pi Now suppose, for the sake of contradiction, that hx2 1,pi = ZxIf we look at the behaviour as x approaches zero from the right, the function looks like this \begin{matrix}x & f(x) = \frac{1}{x^2} \\ 1 & 1 \\ 01 & 100 \\ 001 & \\ 0001 & \\ & \end{matrix}
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X Well begun is half done You have joined No matter what your level You can score higher Check your inbox for more details {{navliveTestEngineeringCount}} Students Enrolled {{navliveTestMedicalCount}} Students Enrolled Start Practicing6041/6431 Spring 08 Quiz 2 Wednesday, April 16, 730 930 PM SOLUTIONS Name Recitation Instructor TA Question Part Misc 6 Let f = {("x, " 𝑥2/(1𝑥2)) x ∈ R } be a function from R into R Determine the range of f f = { ("x , " 𝑥2/(1𝑥2)) x ∈ R } We find different values of 𝑥2/(1 𝑥2) for different values of x Domain Value will always be between 0 & 1 We note that Value of range
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1 f X(y 1 n) where, f X() is the pdf of X which is given Here are some more examples Example 1 Suppose Xfollows the exponential distribution with = 1 If Y = p X nd the pdf of Y Example 2 Let X ˘N(0;1) If Y = eX nd the pdf of Y Note Y it is said to have a lognormal distribution Example 3 Let Xbe a continuous random variable with pdf f⇒ f (x) will be increasing for 1 ≤ x < 2 and will have the least value at x = 1 for the considered interval ⇒ f ( 1 ) = 1 1 2 2 = 1 3 = 3 C a s e 2 − Taking 2 ≤ x < 3 ,> 1 Taylor polynomials > 11 The Taylor polynomial Let f(x) be a given function, for example ex,sinx,log(x) The Taylor polynomial mimics the behavior of f(x) near x= a T(x) ≈f(x), for all x"close" to a Example Find a linear polynomial p 1(x) for which ˆ p 1(a) = f(a), p0 1 (a) = f0(a) p 1 is uniquely given by p 1(x) = f(a)(x−a)f0(a) The graph of y= p
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Transcript Ex 12, 9 Let f N → N be defined by f (n) = { ((𝑛 1)/2 ", if n is odd" @𝑛/2 ", if n is even" )┤ for all n ∈ N State whether the function f is bijective Let f(x)=tanx/x, then log(lim x tends to 0 (f(x)x 2) 1/{f(x)}) is equal to, (where denotes greatest integer function and{} fractional part)=???In mathematics, an injective function (also known as injection, or onetoone function) is a function f that maps distinct elements to distinct elements;
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(a) For any constant k and any number c, lim x→c k = k (b) For any number c, lim x→c x = c THEOREM 1 Let f D → R and let c be an accumulation point of D Then lim x→c f(x)=L if and only if for every sequence {sn} in D such that sn → c, sn 6=c for all n, f(sn) → L Proof Suppose that lim x→c f(x)=LLet {sn} be a sequence in D which converges toc, sn 6=c for all nLet >0To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW Let f(x)=x1x1Then f is differentiable inAnswer to Let f(x) = \dfrac{x^2 3x 2}{x 1} State any inflection points By signing up, you'll get thousands of stepbystep solutions to
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Example \(\PageIndex{1}\) Let the random variable \(X\) denote the time a person waits for an elevator to arrive Suppose the longest one would need to wait for the elevator is 2 minutes, so that the possible values of \(X\) (in minutes) are given by the interval \(0,2\)X!af(x), we can derive many general laws of limits, that help us to calculate limits quickly and easily The following rules apply to any functions f(x) and g(x) and also apply to left and right sided limits Suppose that cis a constant and the limits lim x!a f(x) and lim x!a g(x) exist (meaning they are nite numbers) Then 1lim x!af(x) g(x1 2, but if we let x ˘ 1 n2 for some n ‚k then the left side is at least 1/2, generating a contradiction The other cases follow similarly By Theorem 712, f is continuous on intervals where it converges uniformly, which in this case happens to include the set of sets where it converges pointwise Lastly, we prove that f is not bounded Let
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Let f(x) = e x – x and g(x) = x 2 – x, ∀ x ∈ R Then the set of all x ∈ R where the function h(x) = (fog)(x) is increasing, is A \(\left { 1,\frac{{ 1}}{2}} \right \cup \left {\frac{1}{2},\infty } \right)\)Then , `f (f(x))` = 1 (1 x ) = 2 x ∵ 1 ≤ f (x) < 2 When ,1 < x ≤ 2 Then , f (x) = 1 x Now when , 1 < x ≤ 2 then,2 < x 1 ≤ 3 Then , f (f(x)) = 3 − ( 1 x ) = 2 − x ∵ 2 ≤ f(x)Then X is approximately N(np,np(1 − p)) • Rule of thumb this approximation is reasonably good for np(1 − p) > 10 • P(X = k) ≈ P(k − 1/2 < Y < k 1/2) where Y ∼ N(np,np(1 − p)) • Note P(X≤ k) is usually approximated by P(Y < k 1/2) 25
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F (x) ˘ X1 n˘1 cnI(x¡xn) (a •x •b) converges uniformly, and that f is continuous for every x 6˘xn Solution Let fk(x) ˘ Xk n˘1 cnI(x¡xn) By the Weierstrass Mtest (Theorem 710) with Mn ˘jcnj, {fk(x)} converges uniformly to f (x) Let E ˘ a,b\{xn n 2 N} Since each fk(x) is continuous on E, then by Theorem 712 we know that fTo ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW If` 2f(x)3f(1/x)=x^21` then `f(x)` is2 (a) Define uniform continuity on R for a function f R → R (b) Suppose that f,g R → R are uniformly continuous on R (i) Prove that f g is uniformly continuous on R (ii) Give an example to show that fg need not be uniformly continuous on R Solution • (a) A function f R → R is uniformly continuous if for every ϵ > 0 there exists δ > 0 such that f(x)−f(y) < ϵ for all x
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That is, f(x 1) = f(x 2) implies x 1 = x 2 In other words, every element of the function's codomain is the image of at most one element of its domain 2 Let f (x) = { (1 cos 4x)/x^2, if x < 0 and a, if x = 0 and x/ (√ (16 √x) 4), Let f (x) = { (1 cos 4x)/x2, if x < 0 and a, if x = 0 and √x/ (√(16 √x) 4), if x > 0} If f (x) is continuous at x = 0, determine the value of a Please log in or register to add a comment Let the functioin 'f' be defined by f (x) = 5x² 2 ∀ x ∈ R, then 'f' is (a) onto function (b) oneone, onto function (c) oneone, into function (d) manyone into function Answer Answer (d) manyone into function
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Click here👆to get an answer to your question ️ Let f(x) = x^2 1x^2 and g(x) = x 1x, x ∈ R { 1, 0, 1 } If h(x) = f(x)g(x) , then the local minimum value of h(x) is Join / LoginTimes If the coin lands tails, you roll the die 101 times Let X be 1 if the coin lands heads and 0 if the coin lands tails Let Y be the total number of times that you roll a 6 Find P(X = 1Y = 15) We can find P(X = 1Y = 15) using Bayes' Rule It's easiest using the "odds" form P(X = 1Y = 15) P(X = 0Y = 15) = P(X = 1)F '(x) = lim h→0 1 (xh)2 − 1 x2 h ∴ f '(x) = lim h→0 1 h ⋅ x2 − (x h)2 (x h)2 x2 ∴ f '(x) = lim h→0 x2 −(x2 2hx h2) h(x h)2x2 ∴ f '(x) = lim h→0 x2 − x2 −2hx − h2 h(x
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F (a) 2 (x −a)2 Check that P 2(x) has the same first and second derivative that f (x) does at the point x = a 43 Higher Order Taylor Polynomials We get better and better polynomial approximations by using more derivatives, and getting higher degreed polynomials The Taylor Polynomial of Degree n,forx near a is given by P n(x) = f (a) f In finding the limit, we'll need to consider the left and right limits separately because the rule changes at x = 1 The limit from the right is lim x→1 f (x) = (1)2 = 1 The limit from the left is lim x→1− f (x) = 2(1) = 2 Because the right and left
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