The gradient of the curve at A is equal to the gradient of the curve at B. Given that point A has x coordinate 3, find the x coordinate of point B.

We are told f(x) = (2x-5)^2 (x+3).In part b) we are asked to show that f'(x) = 12x^2 -16x -35, so for part (c) we shall assume this definition for f'(x). We are told that the x coordinate for A is 3. Call the x coordinate for B b. Then, because the gradient at A is equal to the gradient at B, we know that f'(3) = f'(b). From the definition of f'(x) given in part b) we can evaluate f'(3) by plugging in 3 for x. Then 12(3^2) - (16*3) - 35 = 25. So the gradient at A is 25, which means the gradient at B is also 25, i.e. f'(b) = 25. Now that we know this, we can write f'(x) = 12x^2 -16x - 35 = 25, which is a quadratic which we can easily solve.Subtracting 25 from both sides gives us 12x^2 -16x - 60 = 0. Simplifying by dividing each term by a common factor of 4, we obtain 3x^2 - 4x - 15 = 0. (This step is not necessary but may make factorising easier). Now we know that one of the solutions for this quadratic must be x=3, since the x coordinate of A is 3. Therefore (x-3) = 0 is a factor. Factorising fully gives us (x-3)(3x+5) = 0. If students are not confident with this approach, other methods that they are more comfortable with such as the a,b,c method work just as well.So from our factorised equation we can deduce that the two solutions are x=3 and x=-5/3. But x=3 is the coordinate for A, therefore the x coordinate for B is -5/3.