WebObviously the partial derivatives exist at (x,y) = (0,0), as it's a rational function and the denominator is never zero Now to find the partial derivatives at (0,0) you have limx→0 xf
選択した画像 y=x^2 2x-8 domain 147950
What is the domain and range of, y=2x^28 What are the domain and range, y=x^28x16 What are the domain and range, y=2x^212x15 What are the domain and range, y=2x^ Categories Uncategorized Leave a Reply Cancel reply Your email address will not be published Required fields are marked * CommentY(4) = 2 1 Rewriting the LHS in di erential form and factoring the RHS we get dy dx = (x 2)(y 1) (x 3)(y 1) 2 Separating the variables leads to y 1 y 1 dy = x 2 x 3 dx 3 To evaluate the integrals Z y 1 y 1 dy = Z x 2 x 3 dx we need usubstitution on both sides On the LHS, let u = y 1 and then du = dy Given function f(x) = 2x5y = f(x) For x = 0 y = 5 For x = 1 y = 3 For x = 1 y = 7 Substtute x = 0,1,1 respectively, the value of y would be 5,3,7 x values are domain of the function In this case we can substitute any real number for x, we will get value of f(x) Domain of the function is all real numbers Interval notation is (∞,∞)
7 2 Abs Value Function
Y=x^2 2x-8 domain
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