# Stationary points. How to find a stationary point on a graph (turning points)

A stationary point (or turning point) is the point on the graph where the gradient is 0.

That is a stationary point occurs when f´(x) = 0

So to find a stationary point:

**1) Differentiate f(x)
to give f´(x)**

**2) Make f´(x) = 0 and
solve this equation. **

**Example 1**

Find the x coordinate of the stationary point on the graph f(x) = 7x² - 3x + 5

Let’s follow the 2 steps outlined above:

1) Differentiate f(x) to give f´(x)

f´(x) = 14x -3

2) Make f´(x) = 0 and solve this equation.

14x – 3 = 0

14x = 3

x = 3/14

So the turning point of this graph occurs at x = 3/14

**Example 2**

Find the x coordinates of the stationary points on the graph f(x) = x³ - 6.5x² + 12x + 1

Again, to find the stationary points follow the 2 steps above:

1) Differentiate f(x) to give f´(x)

f´(x) = 3x² - 13x + 12

2) Make f´(x) = 0 and solve this equation.

3x² - 13x + 12 = 0

Since you are left with a quadratic it needs to be factorised before it can be solved.

3x² - 13x + 12 = (3x -4)(x-3)

So either 3x-4 = 0 or x -3 = 0

So x = 4/3 or x = 3.

If you draw the graph of f(x) = x³ - 6.5x² + 12x + 1 then you can see that these 2 points are correct.

To verify the nature of a stationary point (either min, max or inflexion) you need to calculate the second derivate. I will write about this in a hub shortly.

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