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JAMB Mathematics 2011 Paper

Question : 34

Total: 46

Find the value of x at the minimum point of the curve y = x3 + x2 - x + 1

Solution:

Concept:

To locate a minimum point, set the first derivative to zero and confirm using the second derivative test.

Explanation:

Given

y = x^3 + x^2 - x + 1

Find the first derivative:

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

At a turning point,

\frac{dy}{dx} = 0

3x^2 + 2x - 1 = 0

Factorise:

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

x = \frac{1}{3}

x = -1

Find the second derivative:

\frac{d^2y}{dx^2} = 6x + 2

Test each value.

For

x = \frac{1}{3}

\frac{d^2y}{dx^2} = 6\left(\frac{1}{3}\right) + 2 = 2 + 2 = 4 > 0

Since

\frac{d^2y}{dx^2} > 0

, this is a minimum point.

For

x = -1

\frac{d^2y}{dx^2} = 6(-1) + 2 = -4 < 0

This is a maximum point.

Answer:

The value of

x

at the minimum point is

\frac{1}{3}

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