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SS3: MATHEMATICS - 2ND TERM

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  1. Matrices I | Week 1
    6 Topics
  2. Matrices II | Week 2
    1 Topic
    |
    1 Quiz
  3. Commercial Arithmetic | Week 3
    7 Topics
    |
    1 Quiz
  4. Coordinate Geometry | Week 4
    8 Topics
    |
    1 Quiz
  5. Differentiation of Algebraic Expressions | Week 5 & 6
    7 Topics
  6. Application of Differentiation | Week 7
    4 Topics
    |
    1 Quiz
  7. Integration | Week 8
    8 Topics
    |
    1 Quiz



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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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