Let f be a twice differentiable function defined on R such that f(0) = 1, f '(0) = 2 and f '(x) ≠ 0 for all x ∈ R asked Mar 3 in Mathematics by Panya01 ( k points) jeeGet the free "Solve f(x)=0" widget for your website, blog, Wordpress, Blogger, or iGoogle Find more Mathematics widgets in WolframAlphaClearly f (x y) = f (x) f (y) for all x, y ==> f (0 0) = f (0)f (0) ==> f (0) = {f (0)}^ (2) ==> f (0) {f (0) 1} = 0 which in turn gives either f (0) = 0 or f (0) = 1 Now ;
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F '(x) 0 and f ''(x) 0 for all x-In this *improvised* video, I show that if is a function such that f(xy) = f(x)f(y) and f'(0) exists, then f must either be e^(cx) or the zero function It'CBSE CBSE (Science) Class 12 Question Papers 1851 Textbook Solutions Important Solutions 4562 Question Bank Solutions Concept Notes & Videos 725 Time Tables 18 Syllabus
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(b) f, h are differentiable at 0, and f′(0) = h′(0) Does it follow that g is differentiable at 0?Ngconverges pointwise to fon 0;1, where f(x) = (1 x= 0 0 0 If f(x) is a quadratic expression such that f(x)0 for all x belongs to R, and if g(x)= f(x) f '(x) f "(x), then prove that g(x)0 for all x belongs to R Share with your friends Share 0 let the quadratic expression f (x) = a x 2 b x c since f (x
In the positive sdirection at s = 0, which is at the point (x0,y0,f(x0,y0)) The directional derivative is denoted Duf(x0,y0), as in the following definition Definition 1 The directional derivative of z = f(x,y) at (x0,y0) in the direction of the unit vector u = hu1,u2i is the derivative of the cross section function (1) at s = 0 Duf(x0,y0{x = 2 f f 2 − 6 f 1 − f 1 ;If f′′(x) > 0 for all x ∈(a,b), then f is concave upward on (a,b) If f ′′ (x) < 0 for all x ∈(a,b), then f is concave down on (a,b) Defn The point (x 0 ,y 0 ) is an inflection point if f is continuous
The Indicator function that is 1 on the irrational numbers and zero elsewhere is Lebesgueintegrable and has integral 1 on 0,1, even though the set on which it is zero the rational numbers in 0,1 is dense The claim may be true if we restrict ourselves to Riemann integration, as the above indicator function is not RiemannintegrableSequences of Functions Uniform convergence 91 Assume that f n → f uniformly on S and that each f n is bounded on S Prove that {f n} is uniformly bounded on S Proof Since f n → f uniformly on S, then given ε = 1, there exists a positive integer n 0 such that as n ≥ n 0, we have f n (x)−f (x) ≤ 1 for all x ∈ S (*) Hence, f (x) is bounded on S by the followingWe just need to find the answer choice for which plugging in x yields the same value as plugging in 1 – x Let's say x is 4 Then the function
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A) If f'(x) >0 on an interval, then f is increasing on that interval b) If f'(x) 0 on an interval, then f is concave upward on that interval d) If f''(x)9 Let f be a continuous real function on R1, of which it is known that f 0(x) exists for all x 6= 0 and that f (x) → 0 as x → 0Dose it follow that f0(0) exists?Graph f (x)=0 f (x) = 0 f ( x) = 0 Rewrite the function as an equation y = 0 y = 0 Use the slopeintercept form to find the slope and yintercept Tap for more steps The slopeintercept form is y = m x b y = m x b, where m m is the slope and b b is the yintercept y = m x b y = m x b Find the values of m m and b b using the
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For example you can prove using some algebra that if f is defined by f(x)=(ax1)(xd) where a^2=d^2=1 and ad0 then f(f(x))=1/x for all real x Choosing a=1,d=1, this give a solution to the problem in C4 Answers4 Hint f can't have a positive maximum at c since then f(c) > 0, f ′ (c) = 0, f ″ (c) ≤ 0 implies that f ″ (c) f ′ (c) − f(c) < 0 Similarly f can't have a negative minimum Hence f = 0 Let x = c be the x coordinate of absolute max of f(x) on a, bIn(x) C because the derivative of In(x) is x None of the listed answers In ()xl) C because we can't take the logarithm of negative numbers In (x) C because the absolute value makes our function become increasing and you can't integrate a
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And then you can get f(x) = 0 for all x ∈ 0, 1 Here is a proof by contradiction Assume that the assertion is not true for some x0 ∈ (0, 1 Thus, f(x0) ≠ 0(f(x0) > 0 ,say) Now consider the interval 0, x0 ,by maximumminimum theorem, f attains its maximum value on 0, x0Show that F(X) = E1/X, X ≠ 0 is a Decreasing Function for All X ≠ 0 ?Example If f(0) = 5, f0(x) exists for all x and 1 f0(x) 3 for all x, show that 5 f(10) 35 4 Mathematical Consequences With the aid of the Mean Value Theorem we can now answer the questions we posed at the beginning of the section
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In other words, we are looking for the xintercept, since y=0 for all xintercepts So we substitute 0 in for f(x) and we get Now we solve for x Add 12 to both sides Divide both sides by 3 This will isolate x So if we let x=4 we should get f(x)=0, in other words, f(4)=0 So lets verify this Check Plug in x=4 works This verifies our answer HSBC244 has shown a nice graph that has derivative #f'(3)=0# Here are couple of graphs of functions that satisfy the requirements, but are not differentiable at #3# #f(x) = abs(x3)5# is shown below graph{y = abs(x3)5 14, 25, 616, 1185} #f(x) = (x3)^(2/3) 5# is shown on the next graphF n(x) = 0 for all x in R Therefore, {f n} converges pointwise to the function f ≡ 0 on R Example 5 Consider the sequence {f n} of functions defined by f n(x) = n2xn for 0 ≤ x ≤ 1 Determine whether {f n} is pointwise convergent Solution First of all, observe that f n(0) = 0 for every n in N So the sequence {f
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Ex 51, 7 Find all points of discontinuity of f, where f is defined by 𝑓(𝑥)={ (𝑥3, 𝑖𝑓 𝑥≤−3@ −2𝑥, 𝑖𝑓−3(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 >0X = 2 f − f 2 − 6 f 1 − f 1 , x = 1, (f = 0 and f ≤ 3 − 2 2 ) or f ≥ 2 2 3 f = 0 Steps Using the Quadratic Formula Steps for Completing the Square
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If 0 < x ≤ 1, then fn(x) = 0 for all n ≥ 1/x, so fn(x) → 0 as n → ∞;Let f (x) > 0 for all x and f' (x) exists for all x If f is the inverse function of h and h' (x) = 11 log x Then f' (x) will be?Image Transcriptionclose Assuming that f(x) is defined for all x0 such that x is a real number, what is the antiderivative of f(x) =x and why?
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= 1 lim {f (h) 1}/h} ·······(1) (note that f (0) has been taken 1, as it can not be 0, because it will make 3 = 0/math and the derivative mathf^{\;'}(x) /math is positive for every mathx/math , then the function mathfAnd if x = 0, then fn(x) = 0 for all n, so fn(x) → 0 also It follows that fn → 0 pointwise on 0,1 This is the case even though maxfn = n → ∞ as n → ∞ Thus, a pointwise convergent sequence of functions need not be bounded, even if it converges to zero Example 55
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Of f(x) at x= a f0(a) = dy dx x=a = lim h!0 f(a h) f(a) h Geometrically This is the slope of the tangent line to y= f(x) at x= a The equation of the tangent line to y= f(x) at the point (a;f(a)) is (from PointSlope Formula) y f(a) = m(x a) We now know that m= f0(a) Derivatives as Functions We can talk about the derivative at any point x If a≠b then ab is nonzero, (ab)/2 is nonzero, and we can construct an ε>0 st ab>ε So if what you quote is a definition in some system then it must be in place of one of the axioms I know Anyway, having solved part d, I think there is an easier way using the hint If f (x)≠x then between themNote We prove a more general exercise as following Suppose that f is continuous on an open interval I containing x
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Where the righthand equation is the result of solving F(x,y,z) = c for z in terms of the independent variables x and y We differentiate the lefthand equation in (12) with respect to the independent variables x and y, using the chain rule and remembering that z = z(x,y) F(x,y,z) = c ⇒ Fx ∂x ∂x Fy ∂y ∂x Fz ∂z ∂x = 0 ⇒ FxShort Solution Steps f ( x ) = 3 x ^ { 2 } \frac { 3 } { x ^ { 2 } } f ( x) = 3 x 2 − x 2 3 To add or subtract expressions, expand them to make their denominators the same Multiply 3x^ {2} times \frac {x^ {2}} {x^ {2}} To add or subtract expressions, expand them to make their denominators the same Multiply 3 x 2 times x 2 x 2This says that either f(x) = 0 or 1 − 2f0(x) = 0 For the first case, we see that f(x) = 0 will solve our original function, since R x 0 0dx = 0 for all x In the second case, f0(x) = 1 2, so f(x) = 1 2 x C To get the value of C, notice in the original equation that if x = 0, then Z 0 0 f(x)dx = (f(0))2 ⇒ f(0) = 0 Thus, C = 0
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NOT continuous at x = 0 Q The function f 2 is 2 continuous at x = 0 and NOT differentiable at x = 0 R The function f 3 is 3 differentiable at x = 0 and its derivative is NOT continuous at x = 0 S The function f 4 is 4 diffferentiable at x = 0 and its derivative is continuous at x = 0Solution • Yes, it does follow that g is differentiable at 0 • Condition(a) implies that f(0) = g(0) = h(0) and therefore also that f(x)−f(0) ≤ g(x)−g(0) ≤ h(x)−h(0) For x > 0, we have f(x)−f(0) x0 and as x ?
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Video Solution For which of the following functions f is f (x) = f (1x) for all x?Homework 5 Solutions Igor Yanovsky (Math 151A TA) Problem 1 Using Taylor expansion, show that f0(x0)= f(x0 h)−f(x0) h − h 2 f00(ξ), for some ξ lying in between x0 and x0 h Solution We expand the function f in a first order Taylor polynomial around x0 f(x)=f(x0)(x− x0)f0(x0)(x−x0)2 f00(ξ) 2 Transcript Ex 51, 8 Find all points of discontinuity of f, where f is defined by 𝑓(𝑥)={ (𝑥/𝑥, 𝑖𝑓 𝑥≠0@&0 , 𝑖𝑓 𝑥=0)┤ Since we need to find continuity at of the function We check continuity for different values of x When x = 0 When x > 0 When x < 0 Case 1 When x = 0 f(x) is continuous at 𝑥 =0 if LHL = RHL = 𝑓(0) Since there are two different
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Ok, so what does f (x) = f (1x) mean?CBSE CBSE (Arts) Class 12 Question Papers 17 Textbook Solutions Important Solutions 24 Question Bank Solutions Concept Notes & Videos 531 Time Tables 18 SyllabusIf F(x)=x has no real solution then also F(F(x)=x has no real solution
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(a) f(x) ≤ g(x) ≤ h(x) for all x ∈ R, and f(0) = h(0);A relative min exists at x = 2, Point (2, 0) but f(x) is not differentiable at x = 2, there is a corner in the graph f '(x) DNE4 Determine from the graph whether f possesses extrema on the interval (a, b) Free Online Scientific Notation Calculator Solve advanced problems in Physics, Mathematics and Engineering Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation History
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0 (b) Explain the shape of the graph by computing the limit as x ?F(x) = (d) Use a graph of f '' to estimate the xcoordinates of the inflection points (Round your Question Consider the function below f(x) = x1/x, x >Show that F(X) = (X − 1) Ex 1 is an Increasing Function for All X > 0 ?
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For all x and y such that −1 < x < 1 (otherwise the denominator f X (x) vanishes) and < < (otherwise the conditional probability degenerates to 0 or 1) One may also treat the conditional probability as a random variable, — a function of the random variable X , namely, If f R → R is a twice differentiable function such that f '' (x) > 0 for all x ∈ R, and f f(1/2) = 1/2, Sarthaks eConnect Largest Online Education Community If f R → R is a twice differentiable function such that f '' (x) > 0 for all x ∈ R, and f f(1/2) = 1/2, f(1) = 1, then Login Remember Register f (x) = x = {x if x ≥ 0 −x if x < 0 So, lim x→0 x = lim x→0 x = 0 and lim x→0− x = lim x→0− ( − x) = 0 Therefore, lim x→0 x = 0 which is, of course equal to f (0) To show that f (x) = x is not differentiable, show that f '(0) = lim h→0 f (0 h) − f (0
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