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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