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Finding the gradient (m): The gradient is the change in y-axis over the change in x-axis Δy = -10-4= -14 Δx = 12-5=7 Δy/Δx = -14/7 = -2 Ac...
f(x) = 3x^3 + 2x^2 - 8x + 4
f(2) = 3(2)^3 + 2(2)^2 - 8(2) + 4
f(2) = 3(8) + 2(4) - 8(2) + 4
f(2) = 24 + 8 - 16 +4
f(2) = 20
a) By simple parametric differentiation of each term, dy/dx = 2x^2 + 2x b) The condition for a turning point is the gradient (dy/dx) at that point is zero. 0 = x(2x + 2) so either x=0 or 2x+2=0. Therefore...
We integrate by parts: u=x u'=1; v'=sin(x) v=-cos(x) integral(xsin(x)dx) = -xcos(x) + integral(cos(x)dx) = -x*cos(x) + sin(x) + C where C is a constant.
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