Let f(x) = 3x^4 - 8x^3 - 3. Find the x- values of the stationary points of this function.
Stationary points occur when f'(x) = 0. To find this, we differentiate f(x) to get f'(x) = 12x^3 - 24x^2. We know that at the stationary points are when f'(x) = 0. so we know that 12x^3 - 24x^2 = 0. We can factorise this to get 12x^2(x - 2) = 0. We can solve this equation to get 12x^2 = 0 and x - 2 = 0. From this we get x = 0 or x = 2. The two x -values of the stationary points of f(x) are 0 and 2.
YS
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