That is, the rule for this transformation is –f (x) To see how this works, take a look at the graph of h(x) = x 2 2x – 3F ix i 2 Evaluate X4 k=1 k2 2 Some examples Example Evaluate X4 r=1 r3 Solution This is the sum of all the r3 terms from r = 1 to r = 4 So we take each value of r, work out r3 in each case, and add the results Therefore X4 r=1 r3 = 13 23 33 43 = 1764 = 100 Example Evaluate X5 n=2 n2 Solution In this example we have used the letter n to represent the variable in the sum, rather0Given that x >

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Free PreAlgebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators stepbystepEx 12, 10 Let A = R − {3} and B = R − {1} Consider the function f A → B defined by f (x) = ((x − 2)/(x − 3)) Is f oneone and onto?3/4 units and run 1, so the slope is 1 &
51 Maxima and Minima A local maximum point on a function is a point (x, y) on the graph of the function whose y coordinate is larger than all other y coordinates on the graph at points close to'' (x, y) More precisely, (x, f(x)) is a local maximum if there is an interval (a, b) with a <đạo hàm trên \(\mathbb{R}\) làOr e x can be defined as f x (1), where f x R → B is the solution to the differential equation df x / dt (t) = x f x (t), with initial condition f x (0) = 1;
Solutions to Graphing Using the First and Second Derivatives SOLUTION 1 The domain of f is all x values Now determine a sign chart for the first derivative, f ' f ' ( x) = 3 x2 6 x = 3 x ( x 2) = 0 for x =0 and x =2 See the adjoining sign chart for the first derivative, f ' Now determine a sign chart for the second derivativeSolution Steps f ( x ) = x ^ { 4 } ( x 1 ) ^ { 3 } f ( x) = x 4 ( x − 1) 3 Use binomial theorem \left (ab\right)^ {3}=a^ {3}3a^ {2}b3ab^ {2}b^ {3} to expand \left (x1\right)^ {3} Use binomial theorem ( a − b) 3 = a 3 − 3 a 2 b 3 a b 2 − b 3 to expand ( x − 1) 3And, as in the first example, =!



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F(x) = ˆ cx4 0 <x<1 0 otherwise Find a E(X) b Var(X) Example 8 To be a winner in the following game, you must be succesful in three succesive rounds The game depends on the value of X, a uniform random variable on (0,1) If X>01, then you are succesful in round 1;Using our notation above, f(x) = x 2 f(1) = 1, f(2) = 4, f(3) = 9, f(4) = 16, f(5) = 25, f(6) = 36 P(X = 1) = 1/6, P(X = 2) = 1/6, etc So E(X 2) = 1/6 4/6 9/6 16/6 25/6 36/6 = 91/6 = The expected value of a constant is just the constant, so for example E(1) = 1 Multiplying a random variable by a constant multiplies the expected value by that constant, so E2X = 2EX ACounselling Sessions to those students who faced difficulties in Maths



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Solution to Algebra I question It f(x) is an exponential function where f(1)=8 and f(85)=87, then find the value of f(3), to the nearest hundreth⃤ PlainExtended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology &It follows that f x (t) = e tx for every t in R Lie algebras




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Explanation Let f (x) = 12xx^34x^5 and note that for every x, x is a root of the equation if and only if x is a zero of f f has at least one real zero (and the equation has at least one real root) f is a polynomial function, so it is continuous at every real number In particular, f is continuous on the closed interval 1,0Professionals For math, science, nutrition, historySafety How works Test new features Press Copyright Contact us Creators



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