Find the turning points of the equation y=4x^3-9x^2+6x?
To find the turning points, first find the first derivative of the equation by differentiating. Differentiating can be done in this equation by multiplying the number in front of the x by the power then lowering the power by one. In the first case, 4x3 would result, 4x3=12 and 3-1=2, so 12x2. Repeat for the other two terms to give dy/dx=12x2-18x+6. Then to find the turning point set dy/dx=0 and solve for x. In this case a factor of 6 can be taken out to give 2x2 -3x+1=0; this can be solved in many ways by factorising, completing the square or using the quadratic formula. Here it would be easiest to use factorising as the brackets must be (2x+a)(x+b) where a =-1 and b=-1, which gives the expansione 2x2 +(a+2b)x+ab and leads to ab=1 and 2b+a=-3, to give (2x-1)(x-1)=0 so the two x values of the turning points are x=1 and x=1/2.
Now to find the y values from the x values, substitute the x values back into the original equation y=4x3-9x2+6x, for x=2 this gives y=4(1)3-9(1)2+6(1)=4-9+6=1. For x =1/2, y=4(1/2)3-9(1/2)2+6(1/2)=4/8-9/4+6/2=5/4. This gives the two values for the turning point are (1, 1) and (1/2, 5/4).
