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Different Steve
Answers (2)
unbalanced parentheses, and an odd function. Is there an arccos somewhere in there? If you meant d/dx 1/cos[(3cosx-2sinx)/√3] d/dx √3 sec(3cosx-2sinx) = √3sec(3cosx-2sinx)tan(3cosx-2sinx) * (-3sinx-2cosx) Not much that can be done with that.
Actually, note that y=-x^2+x-5 so it opens downward y = -(x^2 - x + 1/4) - 5 + 1/4 = -(x - 1/2)^2 - 19/4 Now it should be clear where the vertex lies.
