Lesson 6, Topic 3
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Stationary Points

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A point on a curve at which $$\frac{dy}{dx} = \scriptsize 0$$is called a stationary point and the value of the function represented by the curve at that point is called its stationary value.

At such points the tangent is parallel to the x axis. To find the stationary point, let $$\frac{dy}{dx} = \scriptsize 0$$and solve the resulting equation.

Example:

Find the stationary points of the function

4x3 + 15x2 – 18x + 7

Solution:

y = 4x3 + 15x2 – 18x + 7

$$\frac{dy}{dx}$$ = 12x2 + 30x – 18

At stationary point  $$\frac{dy}{dx} = \scriptsize 0$$

12x2 + 30x – 18 = 0

6(2x2 + 5x – 3) = 0

6(2x – 1)(x + 3) = 0

x = $$\frac{1}{2} \scriptsize\; or \; -3$$

The function has two stationary points and its stationary values (at these points) are given by substituting x = $$\frac{1}{2} \scriptsize \; into \; y$$

y = 4x3 + 15x2 – 18x + 7

$$\scriptsize y = 4(\frac{1}{2})^3 + 15(\frac{1}{2})^2 -18(\frac{1}{2}) + 7$$

= $$\frac{9}{4}$$

substituting x = -3

y = 4(-3)3 + 15(-3)2 – 18(-3) + 7 = 88

$$\frac{9}{4}$$ and 88 are the stationary values of the function 4x3 + 15x2 – 18x + 7

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