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x^4 -8x^2 +15 = 0, we rewrite the equation in square form as (x^2-4)^2 -16 +15 =0 (x^2 -4)^2 = 1 x^2 -4 = ±1 so x^2 = 4±1, (x^2 = 3 or x^2 = 5) Therefore x = {-√3, √3, -√5, √5)

In order to find turning points, we differentiate the function. Hence we get f'(x)=2x + 4. Setting f'(x)=0 we get x = -2 and inputting this into f(x) we get y = 0 therefore the turning point is (-2,0). To...

Let y = arctan(x). Arctan(x) is difficult to differentiate but I know how to differentiate tan(x) (=sec^2(x)) so take the tan of both sides: tan(y) = x. The next step will be to differentiate both sides. ...

6 sinx cosx + (2 sin 2x)/(cos^2 2x)

First we must change our inequality so that we have a zero on one side, In this case we can add three to both sides of the inequality, this gives: x^2 - 6x +5 < 0 Now let's consider the equation y = x^...