Consider the functions f(x) = −x^3 + 2x^2 + 3x and g(x) = −x^3 + 3x^2 − x + 3. (a) Find df/dx (x) and hence show that f(x) has turning points at when x = 2 /3 ± √ 13/ 3 . [5] (b) Find the points where f(x) and g(x) intersect. [4]

a) First differentiate f(x) using standard polynomial derivative rules which gives, df/dx=-3x^2+4x+3. The derivative function gives the gradient of f for any value of x. Turning points occur when the gradient is zero, thus where can find the turning points by finding the values of x that solve df/dx=0. This is -3x^2+4x+3=0. We can use the quadratic formula (-b±√(b^2-4ac))/2a to solve this equation. Substituting in a=-3, b=4, c=3 and using surd rules within the square root gives x = 2 /3 ± √ 13/ 3 as required. b) The functions intersect when they equal one another. i.e. the x coordinates of the points of intersection are found by solving −x^3 + 2x^2 + 3x=−x^3 + 3x^2 − x + 3. Adding x^3 to both sides gives 2x^2 + 3x= 3x^2 − x + 3 which leads to x^2-4x+3=0. This can be factorised as (x-3)(x-1)=0 so the values of x which solve this equation are x=3 and x=1. To find the y coordinates we calculate f(1)=4 and f(3)=0. So the points of intersection are (1,4) and (3,0).