200以上 f(x)=-x 5/2-1/2 ex-2(ex-2-4) 268633

Simplifying F(x) = 25x Multiply F * x xF = 25x Solving xF = 25x Solving for variable 'x' Move all terms containing x to the left, all other terms to the right Add '25x' to each side of the equation 25x xF = 25x 25x Combine like terms 25x 25x = 00 25x xF = 00 Factor out the Greatest Common Factor (GCF), 'x' x(25 F) = 001 solution(s) found fx4x^25 See steps Step by Step Solution Reformatting the input Changes made to your input should not affect the solution (1) "x2" was replaced by "x^2"1) is a critical point The second derivative test f xx = 2;f yy = 2;f xy = 0 shows this a local minimum with

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F(x)=-x 5/2-1/2 ex-2(ex-2-4)-X^2 WolframAlpha Volume of a cylinder?Find the Derivative f(x)=x^(4/5)(x4)^2 Rewrite as Expand using the FOIL Method Tap for more steps Apply the distributive property Apply the distributive property Apply the distributive property Simplify and combine like terms Tap for more steps Simplify each term Tap for more steps Multiply by

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Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!The steps are simple 1 Use the information in the question to work out the value of mathx/math 2 Calculate mathx 3/math Step 1To find equation solutions, solve x1=0 and x1=0 Take the square root of both sides of the equation Subtract 1 from both sides This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, 0 for b, and 1 for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a} Square 0 Multiply 4 times 1

 Let f(x)=x21/x2 and g(x)=x1/x,x∈ R {1,0,1} If h(x)=f(x)/g(x), then the local minimum value of h(x) is (1) 3 (2) – 3 (3) − 2√2 (4) 2√2Δ = b 24ac Δ = 5 24·(1)·(5) Δ = 5 The delta value is higher than zero, soWeekly Subscription $199 USD per week until cancelled Monthly Subscription $699 USD per month until cancelled Annual Subscription $2999 USD per year until cancelled

I assume you mean How can I find F(2) if F (x^2–5x4) = x3?(a) f′′(x) ≤ 0 for x ≥ 0 (b) Since t2/2 is convex we have t2/2 ≥ x2/2x(t−x) = xt−x2/2 This is the general inequality g(t) ≥ g(x)g′(x)(t−x), which holds for any differentiable convex function, applied to g(t) = t2/2 Another (easier?) way to establish t2/2 ≤ −x2/2xt is to note that t2/2x2/2−xt = (1/2)(x−t)2 ≥ 0 Now just move x2/2−xt to the other side Explanation f (x) = {x2 if x ≥ 1 2x if x < 1 f is continuous at 1 if and only if lim x→1 f (x) = f (1) (existence of both is implied) We see from the definition of f above that f (1) = (1)2 = 1 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

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2x25x=3 Two solutions were found x = 3 x = 1/2 = 0500 Rearrange Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation Hint If (this was the original post) $$2f(x) f(1x) = x^2$$ then $$2f(1x) f(x) = (1x)^2$$ So, multiplying the first by $2$ and adding to the second, we have $$3f(x)=2x^2(1x)^2$$All equations of the form a x 2 b x c = 0 can be solved using the quadratic formula 2 a − b ± b 2 − 4 a c The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction x=\frac {\left (3\right)±\sqrt {\left (3\right)^ {2}4\left (1\right)\times 5}} {2\left (1

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Simple and best practice solution for F(x)=(x5)(2x1) equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homeworkAlgebra Find the Properties f (x)= (x5)^2 f (x) = (x − 5)2 f ( x) = ( x 5) 2 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 = 5 h = 5 k = 0 k = 0 Since the value of a a is positive, the parabola opens up Opens UpSimple and best practice solution for f(x)=2(x2)(x4)4 equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework

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F(x)4x25 This deals with factoring multivariable polynomials Overview;Use the distributive property to multiply 3 x 4 − 1 8 x 3 2 4 x 2 by x 1 and combine like terms 3x^{5}15x^{4}6x^{3}24x^{2} 3 x 5 − 1 5 x 4 6 x 3 2 4 x 2Δ = b 24ac Δ = 4 24·(1)·5 Δ = 36 The delta value is higher than zero, so the equation

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At 𝑥 = 2 ?Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyYeah, that's for it That means that F of X is equal to X to the 5th minus X cubed for acts class C Ever We're told that F of 1 equals two So f of negative one Is going to give us right here, this will be a 1 This will be a negative one Supposed to be plus one This will be a negative for some 4

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Piece of cake Unlock StepbyStep Extended Keyboard Examples Integrate f '(x) = ∫f ''(x)dx Please remember the constant f '(x) = −2x− 1 2 C Evaluate at the given initial condition, 4 f '(4) = 2 = − 2(4)− 1 2 C C = 3 Write f' (x) with the value of C f '(x) = 3 − 2x− 1 2 4 Answers4 Active Oldest Votes 3 OK, let's use the hint Let u = x 1 Then f ( x 1) = f ( u) = x 2 − 5 x 3 = ( u − 1) 2 − 5 ( u − 1) 3 = u 2 − 7 u 9 So we just got that f ( u) = u 2 − 7 u 9 But the variable name doesn't matter (it can be u, x

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 Empirically Let us assume that the solution is close to f 0 ( x) = x 3 and let us plug in the given equation f 0 ( x) f 0 ′ ( x) = x 3 3 x 2 ≠ x 3 5 x 2 x 2 But we are off by the polynomial 2 x 2 x 2 which is of a lower degree So let us try f 1 ( x) = x 3 2 x 2, giving F/X2 Directed by Richard Franklin With Bryan Brown, Brian Dennehy, Rachel Ticotin, Joanna Gleason A special effects man helps his girlfriend's ex, a cop, with a sting operation, where the ex gets killed Something's off and he investigates with help from an exGraph f(x)=(x2)^25 Find the properties of the given parabola Tap for more steps Use the vertex form, , to determine the values of , , and Since the value of is positive, the parabola opens up Opens Up Find the vertex Find , the distance from the vertex to the focus Tap for more steps

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Subject to the constraint 2x2 (y 1)2 18 Solution We check for the critical points in the interior f x = 2x;f y = 2(y1) =)(0; x=4,x=0,x=5 >"take out a common factor of "x^2 f(x)=x^2(x^2x) "the factors of which sum to 1 are 5 and 4" rArrf(x)=x^2(x5)(x4) "to find zeros set "f(x)=0 rArrx^2(x5)(x4)=0 "equate each factor to zero and solve for x" x^2=0rArrx=0" with multiplicity 2" x5=0rArrx=5 x4=0rArrx=4Solution for F(x)=x23x5 equation (F)=F23F5 We move all terms to the left (F)(F23F5)=0 We add all the numbers together, and all the variables(F^23F5)F=0 We get rid of parenthesesF^23FF5=0 We add all the numbers together, and all the variables1F^24F5=0 a = 1;

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 Ex 75, 11 Integrate the function 5𝑥/((𝑥 1) (𝑥2− 4) ) We can write the integrand as 5𝑥/((𝑥 1) (𝑥2− 4) ) = 5𝑥/((𝑥 1) (𝑥 − 2We are told that mathf(x^2 1) = x^4 5x^2 3/math Using the substitution mathu = x^2 1/math, we thus have mathf(u) = x^4 (2 3)x^2 (1 3 1F(x) = 2x^2–5x4 2(f(x) = 2(2x^2–5x4 )= 4x^2–10x8 f(2x) = 2(2x)^2–5(2x)4 = 8x^2–10x4 4x^2–10x8 = 8x^2–10x4 8x^2–10x4 4x^210x8=0 4x^2–4 = 0

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Math Input NEW Use textbook math notation to enter your math Try it Ex 52, 9 Prove that the function f given by 𝑓 (𝑥) = 𝑥 – 1, 𝑥 ∈ 𝑅 is not differentiable at x = 1 f(x) = 𝑥−1 = { ((𝑥−1), 𝑥−1≥0@−(𝑥−1), 𝑥−1Algebra Find the Vertex f (x)= (x5)^24 f (x) = (x − 5)2 − 4 f ( x) = ( x 5) 2 4 Rewrite the equation in term of x x and y y f (x) = (x−5)2 −4 f ( x) = ( x 5) 2 4 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 = 5 h = 5

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At 𝑥 = 1 ? Ex 51, 3 Examine the following functions for continuity (b) f (x) = 1/(𝑥 − 5) , x ≠ 5 f (x) = 1/(𝑥 − 5) At x = 5 f (x) = 1/(5 − 5) = 1/0 = ∞ Hence, f(x) is not defined at x = 5 So, we check for continuity at all points except 5 Let c be any real number except 5How can I find F(2) in F x^2–5x4) = x3?

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Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutorWeekly Subscription $199 USD per week until cancelled Monthly Subscription $699 USD per month until cancelled Annual Subscription $2999 USD per year until cancelled Transcript Ex 55, 16 Find the derivative of the function given by f (𝑥) = (1 𝑥) (1 𝑥^2) (1 𝑥^4) (1 𝑥8) and hence find f ′(1) Given 𝑓(𝑥)=(1𝑥)(1𝑥^2 )(1𝑥^4 )(1𝑥^8 )" " Let 𝑦=(1𝑥)(1𝑥^2 )(1𝑥^4 )(1𝑥^8 ) Taking log both sides log 𝑦 = log (1𝑥)(1𝑥^2 )(1𝑥^4 )(1𝑥^8 ) log 𝑦 = log (1𝑥)log⁡(1𝑥^2 )log⁡(1

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X = 2 − f 2 1 2 f 4 f 6 x = 2 f 2 1 2 f 4 f 6 , f ≥ 4 2 − 6 or f ≤ − 4 2 − 6 Steps Using the Quadratic Formula Steps for Completing the SquareTopics Terms and topics; Input the Value of x 2 Input the number of terms 5 the sum = Number of terms = 5 value of x =

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Derivative of f (x)=34x^2, x=5 \square!Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!Homework 5 Solutions 3132 f(x;y)=œ xy(x2−y2) x2y2 (x;y)≠(0;0) 0 (x;y)=(0;0) Note fis continuous, (by computing lim(x;y)→(0;0) of the formula above, eg using polar coorinates) (a) Find f x and f y when (x;y)≠(0;0) Away from (0;0);fcan be di erentiated using the formula de ning it,

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Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us! Ex 51, 5 Is the function f defined by 𝑓(𝑥)={ (𝑥, 𝑖𝑓 𝑥≤1@&5, 𝑖𝑓 𝑥>1)┤ continuous at 𝑥 = 0 ?Solution for F(x)=x26x5 equation (F)=F26F5 We move all terms to the left (F)(F26F5)=0 We add all the numbers together, and all the variables(F^26F5)F=0 We get rid of parenthesesF^26FF5=0 We add all the numbers together, and all the variables1F^25F5=0 a = 1;

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Given 𝑓(𝑥)={ (𝑥, 𝑖𝑓 𝑥≤1@&5, 𝑖𝑓 𝑥>1)┤ At x = 0 For x = 0, f(x) = x Since this a polynomial It is continuous ∴ f(x) is continuous for x = 0 At x = 1 f(x) is continuousIf f^2(x) refers to f(x)^2 then f^2(x) = 4x 15 —> f(x) = sqrt(4x 15) If f^2(x) refers to f(f(x)) then suppose f(x) = ax b —> f(f(x)) = af(x) b —> f(f

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