Search over 10,000 free study notes

Stationary points are points where the gradient of a function is equal to 0. In this question the product rule can be used to find the gradient of the given function. The product rule is given by u'v+uv'...

(1 mark) Differentiate equation in the question: 4x³+8x²-4x-8(1 mark) Equate this to zero: (x-1)(x+1)(x+2)=0(1 mark) Find turning points (roots of above equation): x=1,-1,-2(1 mark) ...

Stationary points are found by considering the points at which the gradient of the function equal zero. For the above, you need to employ the quotient rule, since both numerator and denominator are f(x), ...

Set out information given in question, and taking the upward direction to be positive: s (displacement) = ?, u (initial speed) = 21ms^-1, v (final speed at maximum height) = 0ms^-1, a ...

Write y=x²-4x+2k. And m:= (x_m, y_m) for the coordinates of our vertex. We deduce that x_m  is exactly the value of x for which y'=2x-4=0, because m is a minimum p...