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(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
dy/dx=12x^2+6 d^2y/dx^2=24x
When finding the equation of a straight line there are two important figures to calculate. The first being the gradient (the slope of the line) and the second being the y intercept (where the line crosses...
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