Turning Points In A Level Maths
Turning Points – Introduction
The concept of turning points is a key topic in A Level mathematics, specifically in the area of differentiation. Turning points occur when the curve of a function changes direction. There are two types of turning points: maximum points and minimum points.
Maximum Turning Points:
A maximum turning point is a point on a curve where the function reaches its highest value in a specific interval. At a maximum turning point, the curve changes from increasing to decreasing. This means that the slope of the curve changes from positive to negative.
Minimum Turning Points:
On the other hand, a minimum turning point is a point on a curve where the function reaches its lowest value in a specific interval. At a minimum turning point, the curve changes from decreasing to increasing. This means that the slope of the curve changes from negative to positive.
Points of Inflection:
In addition to maximum and minimum turning points, there is another type of turning point called a point of inflection. A point of inflection is a point on a curve where the concavity changes. At a point of inflection, the curve changes from being concave upwards to concave downwards, or vice versa. This means that the second derivative of the function changes sign.
Turning Points - Example 1
Solution
\begin{aligned} & y=x^3-3 x^2+3 x \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^2-6 x+3 \end{aligned}Remember that turning points take place when the gradient function is equal to 0. Putting this expression equal to zero generally means that you now have to solve a quadratic equation which will give you two values for x. You need to test both values.
Also when or if you are testing the gradient either side of a turning point and let us say that one value is positive. It would be wrong to make the assumption that the second value would be negative. You need to still carry out the calculation as the gradient value could be the same either side which then results in a point of inflection.
When 3 x^2-6 x+3=0
\begin{aligned} 3\left(x^2-2 x+1\right) & =0 \\ 3(x-1)^2 & =0 \\ x & =1 \end{aligned}This means x=1, y=1
Now consider points close to (1,1)
\begin{array}{cccc} x & 0.9 & 1 & 1.1 \\ \frac{\mathrm{d} y}{\mathrm{~d} x} & 0.03 & 0 & 0.03 \end{array}You can see that the gradient either side of the turning point is positive which means that there is point of inflection and a sketch of this is shown below:
Turning Points - Example 2
When 4 x^3+9 x^2-10 x-3=0 and using the factor theorem to find that x-1 and hence using long division we obtain the following factorised form or f(x):
\begin{aligned} & (x-1)\left(4 x^2+13 x+3\right)=0 \\ & (x-1)(4 x+1)(x+3)=0 \end{aligned}Solutions are therefore:
x=1, x=-\frac{1}{4} \text { or } x=-3When x = 1:
\begin{aligned} y & =(1)^4+3(1)^3-5(1)^2-3(1)+1 \\ & =-3 \end{aligned}
When x=-\frac{1}{4}
\begin{aligned} y & =\left(-\frac{1}{4}\right)^4+3\left(-\frac{1}{4}\right)^3-5\left(-\frac{1}{4}\right)^2-3\left(-\frac{1}{4}\right)+1 \\ & =\frac{357}{256} \end{aligned}
When x=-3,
\begin{aligned} y & =(-3)^4+3(-3)^3-5(-3)^2-3(-3)+1 \\ & =-35 \end{aligned}
The turning points are therefore:
(1,-3),(-3,-35) \text { and }\left(-\frac{1}{4}, \frac{357}{256}\right)
Turning Points – Exam Style Question
Solution
\begin{aligned} y & =2 x+9 x-24 x+13 \\ \frac{d y}{d x} & =6 x^2+18 x-24 \end{aligned}\begin{aligned} & 6 x^2+18 x-24=0 \\ & x^2+3 x-4=0 \\ & (x+4)(x-1)=0 \\ & x=-4 \quad x=1 \end{aligned}Test using the second derivative when x = – 4 and x = 1:
\begin{aligned} & \frac{d^2 y}{d x^2}=12 x+18 \\ & x=-4 \quad \frac{d^2 y}{d x^2}=12(-4)+18=-30 \end{aligned}So this is a MAX
x=1 \quad \frac{d^2 y}{d x^2}=12(1)+18=30So this is a MIN
So (- 4, 125) is the Max
In summary, turning points of a curve refer to the points where the curve changes direction. Maximum turning points occur where the curve changes from increasing to decreasing, while minimum turning points occur where the curve changes from decreasing to increasing. Points of inflection, on the other hand, indicate a change in concavity.