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Elsewhere Show that Y = X2 has an exponential distribution with mean = 2 Solution Note that the function y= x2 is strictly increasing and hence invertible on the e ective domain with x>0 , and its inverse is given by x= h(y) = p y Then h0(y) = 1 2 p y By the method of transformation, we get the pdf of Y by f Y (y) = f X(h(y)) jh0(y)j = 8 < p ye (pExercise 2 Let f R → R that satisfies the multiplicative property f(x y) = f(x)f(y) for all x,y ∈ R Assume f is not identically equal to zero (i) Show that f(0) = 1 and that f(−x) = 1 f(x) for all x ∈ R Show that f(x) > 0 for all x ∈ R First let us prove two simple Lemmas Lemma 1 f(0) 6= 0 Proof If x/yy/x=1(where x ,y is not equal to zero),then the value of x3y3 is 2 See answers Advertisement Advertisement sivaprasath sivaprasath Answer 0 Stepbystep explanation Given To find the value of if,, x ≠ 0 , y ≠ 0 Solution We know that,

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If y=f(x)=x 2/x-1 then show that x=f(y)-If x = 0, y =0 Stepbystep explanation Hi, to obtain the values of the variable y in each case, we have to replace the variable xFor all x1,x2 ∈ X Show that if X is complete, then Y must also be complete A function f X → Y is a bijection if it is onetoone and onto Equivalently, a function f X → Y is a bijection if it has an inverse f−1 Y → X Solution Let {yn} be a Cauchy sequence in Y

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 Ex 32, 13 If F (x) = 8(cos⁡𝑥&〖−sin〗⁡𝑥&0@sin⁡𝑥&cos⁡𝑥&0@0&0&1) , Show that F(x) F(y) = F(x y) We need to show F(x) F(y) = F(x y) Taking LHS Given F(x) = 8(cos⁡𝑥&〖−sin〗⁡𝑥&0@sin⁡𝑥&cos⁡𝑥&0@0&0&1) Finding F(y) Replacing x by y in F(x) F(y) = 8(cos⁡𝑦&〖−sin〗⁡𝑦&0@sin⁡𝑦&coSOLUTION If f(x)= x2/x1, then f(n1) is equal to A)1/2 b)n2/n1 c)n1/n2 d)n1/n2 Answer by richard1234(7193) (Show Source) You can put this solution on YOUR website!If f(x) = (x 2 2x 1) 2/3, then f'(0) =?

Given >0, pick = Then if d(x;y) < , we have d(f(x);f(y)) cd(x;y)To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW If `2f (x1)f((1x)/x)=x , then f(x)` is 1 Follow 0 Anuradha Sharma, Meritnation Expert added an answer, on 23/7/15 Anuradha Sharma answered this f x = x x 1 = 1 y y = x 1 x xy = x 1 x y 1 = 1 x = 1 1 y So f y = 1 y = 1 x

We show that f is injective To this end, let x 1;x 2 2A and suppose that f(x 1) = f(x 2) Then (g f)(x 1) = g(f(x 1)) = g(f(x 2)) = (g f)(x 2) But since g f is injective f^(1)(x)=sqrt(x1) Let y=f(x) Then, y=x^21 To determine the inverse function, switch the places of x and y, and subsequently solve for y x=y^21 y^2=x1 y=sqrt(x1) Since this is the inverse, y=f^(1)(x) f^(1)(x)=sqrt(x1)If xy=2 and x−y=1, then what is xy?

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I got B but I feel as though it's a trick Here's myProblem1(WR Ch 5 #1) Let f be defined for all real x, and suppose that jf (x)¡ f (y)j•(x¡y)2 for all real x and y Prove that f is constant Solution For x 6˘y, from the above inequality we have jf (x)¡f (y)j jx¡yj •jx¡yj So then jf 0(y)j˘ fl fl fl fllim x!y f (x)¡ f (y) x¡y fl fl fl˘ lim x!y fl fl fl f (x)¡ f (y) xClick here👆to get an answer to your question ️ If x = at^2 and y = 2at , then find d^2ydx^2 Join / Login > 12th > Maths > Continuity and Differentiability > Derivatives of Implicit Functions > If x = at^2 and y = 2at , t maths If x = a t 2 and y = 2 a t,

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2 (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 xX(0)=1 Show that x(t)>t2 for all t for which x is defined There is an easy way to solve this problem, and a harder way Solution 1Definition 11 Let f X → Y be a function We say f is onetoone, or injective, if and only if for all x1,x2 ∈ X, if f(x1) = f(x2) then x1 = x2 Or equivalently, if x1 6= x2, then f(x1) 6= f(x2) Symbolically, f X → Y is injective ⇐⇒ ∀x1,x2 ∈ X,f(x1) = f(x2) → x1 = x2 To show that a function is onetoone when the domain is a finite set

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 Consider f R → 9, ∞) given by f(x) = 5x 2 6x 9 Prove that f is invertible with f1 (y) = (√ 545y3/5) where R is the set of all nonnegative real numbers Example 12 Show that f N → N, given by f(x) = { (𝑥1 , 𝑖𝑓 𝑥 𝑖𝑠 𝑜𝑑𝑑@𝑥−1, 𝑖𝑓 𝑥 𝑖𝑠 𝑒𝑣𝑒𝑛)┤ is both oneone and onto Check oneone There can be 3 cases x1 & x2 both are odd x1 & x2 both are even x1 is odd & x2 is even If x1 & x2 are both odd f(x1) = x1 1 f(x2) = x2 1 Rough OneDivide 0 0 by 4 4 Multiply − 1 1 by 0 0 Add − 1 1 and 0 0 Substitute the values of a a, d d, and e e into the vertex form a ( x d) 2 e a ( x d) 2 e Set y y equal to the new right side Use the vertex form, y = a ( x − h) 2 k y = a ( x h) 2 k, to determine the values of a a, h h, and k k

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 If y=f(x)=x2/x1, x,y is not=1 then x is equal to 1 See answer Advertisement Advertisement Snehapareek975 is waiting for your help Add your answer and earn points rahman786khalilu rahman786khalilu Stepbystep explanation hope it helps mark as brainliest New questions in MathIf x = −1, y =1 ;Now, sin(y)=0 when y=nˇ, for all integers n Then setting 1 xcos(nˇ) =1 (−1)nx=0, we get that critical points are (x;y)=((−1)n1;nˇ) for n∈Z At these points f xx≡0;

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(b) f(x,y) = 2x2 xy2 −2, D= {(x,y) ∈ R2;x2 y2 64} Solution (a) Calculating the partial derivatives gives f x = y 2 √ x − 1, f y = √ x−2y 6 Finding critical points f x = 0, f y = 0 =⇒ (x,y) = (4,4) Note that (4,4) ∈ Dand that f(4,4) = 12 The boundary of Dconsists ofX y = 2 (equation 1) x y = 1 (equation 2) Add the equations for 2x = 3, meaning that x = 3/2 Using x = 3/2 in equation 1 gives y = 1/2 Using x = 3/2 in equation 2 also gives y = 1/2 (3/2)*(1/2) = 3/4 AnswerIf y = (sin^1x)^2 , then show that (1 x^2) d^2ydx^2 x dydx = 2 > 12th > Maths > Continuity and Differentiability > Derivatives of Composite Functions and Chain Rule > If y = (sin^1x)^2 , then

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F yy=−xsin(y)S((−1)n1;nˇ) =−(−1)n1 sin(nˇ)=0 So the Hessian matrix at ((−1)n1;nˇ) is 0 (−1)n (−1)n 0 with determinant D=−(−1)2n =−10 otherwise Now it only remains to nd E X 2 j Y = 2 and E X j Y = 2 Applying integration by parts twice we have E X 2 j Y = 2 = 1 2 x 2 6 e 2 x 6 dx = x 2 e 2 x 6 1 2 12 xe 2 x 6 1 2 12 6 e 2 x 6 1 2 = = 100 On the other hand, again applying4 Suppose that x is a solution to the initial value problem x0=x t2 2t;

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F(x) = (xe x 2 2;Problem 5 A realvalued function f X!R on a metric space Xis lower semicontinuous if f(x) liminf n!1 f(x n) for every x2Xand every sequence (x n) in Xsuch that x n!xas n!11 0 f(x)e sxdx Thus if Xis a continuous random ariablve with the PDF such that f X(x) = 0 for x

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A) 4/3 B) 0 C) 2/3 D) 4/3 E) 2 can someone check my answer?Y=f (x) The y is to be multiplied by 1 This makes the translation to be "reflect about the xaxis" while leaving the xcoordinates alone y=f (2x) The 2 is multiplied rather than added, so it is a scaling instead of a shifting The 2 is grouped with the x, so it is a horizontal scalingTo ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW If `f(x)=(x1)/(x1)` then `f(2x)` is equal to

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(a) f(x;y) = p y x Solution y x 0 is the domain This consists of all points in the plane above or on the line y = x This is a closed set The range of the function is 0;1) Here is a sketch of the level curves f = 0, f = 1, and f = 2 The curves are the lines y x = 0, y x = 1, and y x = 4, respectively (b) f(x;y) = y=x2 Solution x2 6= 0Differentiating the given equation 1 with respect to x, d y / d x = f ′ (2 x − 1 / x 2 − 1) ∗ d / d x (2 x − 1 / x 2 − 1), by chain rule of derivatives From there, simply substitute the value of f'(x) and differentiate the inner quantity and you will end up with the value of dy/dx(a) Show that if g f is injective then f is injective (b) Show that if g f is surjective then g is surjective Solution First, we prove (a) Suppose that g f is injective;

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(2) Show that f R→Rgiven by f (x) =x2 is NOT Lipschitz That is, for any L >0, find x,y ∈Rsuchthat f (x)−f (y)>Lx −y Answer Take x =2L and y =L Thenf (x)−f (y)=3L2 >L2 =Lx −y (3) Provethefollowing if fR→Rand gR→RareLipschitzfunctions,thenthecomposition f gR→Ris Lipschitz Answer Let Lf and Lg be theJust replace x with n1 to obtain so choice C is correct1 y = 1 3 e y 3 for y > 0 Then we have fX jY =2 (x j 2) = (1 6 e 2 x 6 if x > 2;

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4 Definition A literal is any Boolean variable x or its complement x' Truth Tables of Boolean functions Much like the truth tables for logical propositions If f(x,y,z, ) is an nvariable Boolean function, a truth table for f is a table of n1 columns (oneDivide \left(f6\right), the coefficient of the x term, by 2 to get \frac{f}{2}3 Then add the square of \frac{f}{2}3 to both sides of the equation This stepE−x2 −x 357 Show that the function f(X) = X−1 is matrix convex on Sn Solution We must show that for arbitrary v ∈ Rn, the function g(X) = vTX−1v is convex in X on Sn This follows from example 34 41 Consider the optimization problem minimize f0(x1,x2) subject to 2x1 x2 ≥ 1 x1 3x2 ≥ 1 x1 ≥ 0, x2 ≥ 0 Make a

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1 Let f be defined for all real x, and suppose that f(x)−f(y) ≤ (x−y)2 for all real x and y Prove that f is constant Proof f(x)−f(y) ≤ (x−y)2 for all real x and y Fix y, f(x)−f(y) x−y ≤ x−y Let x → y, therefore, 0 ≤ lim x→y f(x)−f(y) x−y ≤ lim x→y x−y = 0 It implies that (f(x) − f(y))/(x − y) → 0 as x → y2 for twice differentiable functions, show ∇2f(x) 0 3 show that f is obtained from simple convex functions by operations that preserve convexity • nonnegative weighted sum • composition with affine function • pointwise maximum and supremum • composition • minimization • perspective Convex functions 3–13Show that it may not be the case if µ(X) = ∞ Solution 1 Notice that ∩ Then f M(x) ∈ L1, and, by Theorem 226, there exists a sequence of continuous functions that converges to f M in L1 By Corollary 232, a subsequence of that sequence converges to f M ae We conclude that there exists a sequence of contin

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Hence, (sn) is a Cauchy sequence and must converge to an element in X 2 Continuous function Question f (X,τX) → (Y,τY) is continuous ⇔ ∀x0 ∈ X and any neighborhood V of f(x0), there is a neighborhood U of x0 such that f(U) ⊂ V Proof "⇒" Let x0 ∈ X f(x0) ∈ Y For each open set V containing f(x0), since f isThe Function which squares a number and adds on a 3, can be written as f (x) = x2 5 The same notion may also be used to show how a function affects particular values Example f (4) = 4 2 5 =21, f (10) = (10) 2 5 = 105 or alternatively f x → x2 5 The phrase "y is a function of x" means that the value of y depends upon the value of All that is required here is to substitute x = 2x into the right side in the same way that is done for a numeric value #f(x)=x^2x# f(2) means substitute x = 2

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1 Example 1 f(x) = x We'll find the derivative of the function f(x) = x1 To do this we will use the formula f (x) = lim f(x 0 0) Δx→0 Δx Graphically, we will be finding the slope of the tangent line at at an arbitrary point (x 0, 1 x 1 0) on the graph of y = x (The graph of y = x 1 is a hyperbola in the same way that the graph of Answer If x = −2, y =8 ;X−y2 − x 6y, D= {(x,y) ∈ R2;0 6x69,0 6y65};

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Basically, the function has to be "onetoone" and "onto" The absolute value function isn't onetoone, because although every x in the domain maps to only one y in the range, not every y in the range is mapped to by only one x But IF the function is invertible, then f (x)=f (a) will imply x=aStack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange If y = sin^1 x/√(1 x^2), then show that (1 x^2)d^2y/dx^2 3xdy/dx y = 0 asked in Mathematics by Samantha ( 3k points) continuity and differntiability

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Verify that f (x,y,z) = 1 p x2 y2 z2 satisfies the Laplace equation f xx f yy f zz = 0 Solution Recall f x = −x/ x2 y2 z2)3/2 Then, f xx = − 1 x2 y2 z2)3/2 3 2 2x2 x2 y2 z2)5/2 Denote r = p x2 y2 z2, then f xx = − 1 r3 3x2 r5 Analogously, f yy = − 1 r3 3y 2 r 5, and f zz = − 1 r3 3z r Then, fGraph f (x)= (x2)^21 f (x) = (x 2)2 − 1 f ( x) = ( x 2) 2 1 Find the properties of the given parabola Tap for more steps Use the vertex form, y = a ( x − h) 2 k y = a ( x h) 2 k, to determine the values of a a, h h, and k k a = 1 a = 1 h = − 2 h = 2 k = − 1 k = 1 Since the value of a a is positive, the

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