Considering the following trinomial function : u(x)=2x²-3x+1 ; Determine the domain of definition of f=√u and make the variation table of f=√u.
example of a variation table : http://homeomath.imingo.net/document/tabvar25.gif
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You need to remember that domain of the function contains the values of x that makes valid the equation of the function.
Notice that the equation of the function `f(x) = sqrt(2x^2 - 3x + 1)` is valid if `2x^2 - 3x + 1 >= 0` , hence, you need to solve the quadratic inequality to find the domain of definition of the function.
You need to solve the attached equation `2x^2 - 3x + 1 = 0` , hence, you may use quadratic formula, such that:
`x_(1,2) = (3+-sqrt(9 - 8))/4 => x_(1,2) = (3+-1)/4 => x_1 = 1 ; x_2 = 1/2`
You need to notice that the expression is positive over intervals `(-oo,1/2]U[1,oo)` , hence, the function increases over `(-oo,1/2]U[1,oo).`
Hence, evaluating the domain of definition of the function yields `x in (-oo,1/2]U[1,oo)` and checking the variation of the function yields that the function increases over `(-oo,1/2]U[1,oo).`
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