Find f(g(x)) f(x)=x^29 , g(x)=3x, Set up the composite result function Evaluate by substituting in the value of into Simplify each term Tap for more steps 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 Answerf(x) = x^4 2x^3 2x^2 x 1 g(x) = x^2 2x 3 Here we get the remainder x 2 when we divide f(x) with g(x) hence ,r(x) must be added to the f(x) to make g(x) exactly divisible by p(x)x 2 = x 2 x 2 should be added New questions in Math are yar koi to ban jao aaga samaj jao itz me cute boy heer pls helo retti or zarra koi to bano where are you in whichLearn how to solve f(g(x)) by replacing the x found in the outside function f(x) by g(x)
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See an explanation below (f g)(x) = f(x) g(x) = x^2 10 x 3 (f g)(x) = x^2 x 10 3 (f g)(x) = x^2 x 7 Algebra ScienceF(x)=x4−9 g(x)=x39 (f⋅g)(x)= Thanks so much!Short 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 2
The Derivative tells us the slope of a function at any point There are rules we can follow to find many derivatives For example The slope of a constant value (like 3) is always 0; Transcript Ex 23, 5 Find the range of each of the following functions f(x) = 2 – 3x, x ∈ R, x > 0 Given that x > 0, Multiplying 3 both sides 3x > 0 × 3 3x > 0 Multiplying 1 both sides – 1 × 3x < – 1 × 0 – 3x < 0 Adding 2 both sides 2 – 3x < 2 0 (We need to make it in form 2 – 3x) 2 – 3x < 2 f(x) < 2 We note that value of f(x) is less than 2 (not including 2) Hence Given 𝑓(x) = 2x − 3 and g(x) = x3 5 𝑓∘𝑔 = f(g(x)) = f(x3 5) = 2(x3 5) – 3 = 2x3 10 – 3 = 2x3 7 ∴ 𝒇∘𝒈 = 2x3 7 Now, we need to check if 𝑓∘𝑔 is invertible and find its inverse We check if it is invertible by checking oneone and onto Let p(x) = 𝒇∘𝒈 = 2x3 7 Checking oneone p(x1) = 2(x1)3 7 p(x2) = 2(x2)3 7 Putting p(x1) = p(x2) 2(x1)3
F g (5) means substitute 5 for x in the right side of g (x), simplify, then substitute what you get for x in the right side of f (x Substitute the equivalents for f (x) and g(x) (f −g)(x) = x2 − 1 − (2x −3) Distribute the minus sign (f −g)(x) = x2 − 1 − 2x 3 Combine like terms (f −g)(x) = x2 − 2x 2 Evaluate at x = 3 (f −g)(3) = 32 − 2(3) 2 (f −g)(3) = 5F (x) = x g' (x) and f' (x) = x g (x) and f (0) = 1/2 and g (0) = 1/3 WolframAlpha Volume of a cylinder?
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Find g (f (x)) f (x)=x^27 , g (x)=3x3 f (x) = x2 − 7 f ( x) = x 2 7 , g(x) = 3x − 3 g ( x) = 3 x 3 Set up the composite result function g(f (x)) g ( f ( x)) Evaluate g(f (x)) g ( f ( x)) by substituting in the value of f f into g g g(x2 − 7) = 3(x2 − 7)−3 g ( x 2 7) = 3 ( x 2 7) 3 Simplify each term 1 What is (f−g)(x)?Free pre calculus calculator Solve precalculus problems stepbystep
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F (x) = 3x1, g (x) = x^2 The Gecko Makes An Edit 15 GEICO Insurance Watch later Share Copy link Info Shopping Tap to unmute If playback doesn't begin shortly, try restarting a^33a^2 Given f(x)=x^2,g(x)=x3 (f*g)(x)=x^2(x3) =x^33x^2 (f*g)(a)=(a)^33(a)^2 =a^33a^2 For two given functions f(x) and g(x), the function (f g)(x) is (f g)(x) = f(x) g(x) Here we have the functions f(x) = x^2 1 g(x) = 5 x Now we want to find (f g)(x), we can use the above expression (f g)(x) = f(x) g(x) = (x^2 1) (5 x) (f g)(x) = x^2 1 5 x = x^2 x 6 (f g)(x) = x^2 x 6
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Alors le calcul est soit la fonction F (x)= (x2) (x1) (x3) si f= 5 Si je reprends les bases ont doit développer (ab) (ab)= ( a − b) 2 et ( a b) 2 =ab Donc, on développe les 2 premières, ensuite on prend la somme entre parenthèse de celleci et on termine le développement avec la dernière parenthèse f'(x) = 12(x 3 x 2 5x7) Cette expression est encore factorisable (heureusement) en remarquant que 1 est une racine évidente du polynôme, ce qui veut dire que f'(1) = 0 Comme cela a été rappelé cidessus, si f'(1) = 0 alors l'expression de f'(x) est factorisable par (x1) (C'est la base de la factorisation des formes polynomiales) Ainsi, f'(x) peut se mettre sous la forme (x1)(ax²Answer by Edwin McCravy () ( Show Source ) You can put this solution on YOUR website!
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The answer is obviously 3 The domain of g is unrestricted so the only caveat is whether f(x) exists for all values of x Some might argue that g(x) exists even if x doesn't exist but others would argue that since g is a mapping from the domain to Suppose I write this another way Let u = x − 1 2 ← g(x) Given that f (x) = 2x 1 then f (u) = 2u 1 But u = x − 1 2 So f (u) = 2 x −1 2 1 '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ f (u) is exactly the same process as f (g(x)) f (g(x)) = 2 x − 1 2In mathematics, even functions and odd functions are functions which satisfy particular symmetry relations, with respect to taking additive inversesThey are important in many areas of mathematical analysis, especially the theory of power series and Fourier seriesThey are named for the parity of the powers of the power functions which satisfy each condition the function () = is
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