If y = x³ - x² - x + 6, find the values of x at the turning point.

Correct Answer is : 1,−131, -1/31,−31

Correct Answer is : 1,−131, -1/31,−31

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WAEC Further Mathematics 2012 Paper

Question : 13

Total: 40

y = x^{3} - x^{2} - x + 6

, find the values of x at the turning point.

Solution:

Concept:

At a turning point, the gradient of the curve is equal to zero.

Explanation:

Given:

y = x^{3} - x^{2} - x + 6

Differentiate with respect to

x

\frac{dy}{dx} = 3x^{2} - 2x - 1

At the turning point, set

\frac{dy}{dx} = 0

3x^{2} - 2x - 1 = 0

Factorise the quadratic:

3x^{2} - 3x + x - 1 = 0

(3x + 1)(x - 1) = 0

Equate each factor to zero:

3x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{3}

x - 1 = 0 \quad \Rightarrow \quad x = 1

Therefore, the turning points occur at

x = -\frac{1}{3}

and

x = 1

Answer:

x = -\frac{1}{3}, 1

This corresponds to option C.

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