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With equations like this the most important thing is that you end up with all the numbers on one hand side and all the letters on the other side. People often think there is a 'right' and 'wrong' order to...
(1.) (a.) f’(x)=3x^2+6x-2
(b.) x=6 gradient=142
(c.) since f’(x)>0 at x=6, the function is increasing.
In order to prove that one real root of an equation is situated in a certain interval, we calculate the value of the function at the ends of the given interval. In the given case, f(-2) = (-2)^3 - 3*(-2) ...
completing the square: (x+2)^2-9 finding the roots: (x+2)^2=9 x+2=+/-3 at y= 0, x= -5 or 1
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