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In order to solve this formula, you would need to collect similar terms to the same side and then simplify to work out the value of 'x'. We can do this by collecting all the 'x' terms to one side first, t...
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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