Asked by GNTC
If the domain of a function F is the set of all real numbers and the domain of a function G is the set of all real numbers, under what circumstances do (F+G)(x) and (F/G)(x) have different domains?
Whenever G(x) = 0 then the domains would differ. For example, if F(x) =x+2 and G(x) = x-3 then (F+G)(x) = (x+2)+(x-3) the domain for (F+G)(x) would be all real numbers. (F/G)(x)= (x+2)/(x-3) would not exist when x=3 because you cannot divide a function by zero.
Whenever G(x) = 0 then the domains would differ. For example, if F(x) =x+2 and G(x) = x-3 then (F+G)(x) = (x+2)+(x-3) the domain for (F+G)(x) would be all real numbers. (F/G)(x)= (x+2)/(x-3) would not exist when x=3 because you cannot divide a function by zero.
Answers
Answered by
Reiny
Whatever the domains of F(x) and G(x) are they would be the same for (F+G)(x)
but for (F/G)(x) you would have to restrict any values for which G(x) = 0
In effect you stated that with an example
but for (F/G)(x) you would have to restrict any values for which G(x) = 0
In effect you stated that with an example
Answered by
Molly
If a graph of a quadratic function can have 0, 1 or 2 x-intercepts. How can you predict the number of x-intercepts without drawing the graph or (completely) solving the related equation?
Suppose that the graph of f(x) = ax^2+bx+c has x-intercepts (m,0) and (n,0). What are the x-intercepts of g(x) = –ax^2–bx–c?
I really don't understand this, any elaboration will be greatly appreciated. thank in advance
Suppose that the graph of f(x) = ax^2+bx+c has x-intercepts (m,0) and (n,0). What are the x-intercepts of g(x) = –ax^2–bx–c?
I really don't understand this, any elaboration will be greatly appreciated. thank in advance
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