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WAEC Further Mathematics 2018 Paper

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Question : 24

Total: 40

Given that

y = x(x+1)^2

, calculate the maximum value of y.

-2
0
1
2

Solution:

Concept:

For a function

y = f(x)

, a maximum occurs where

\frac{dy}{dx} = 0

and

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

.

Explanation:

Expand the given expression:

y = x(x+1)^2 = x(x^2+2x+1) = x^3+2x^2+x

Differentiate with respect to

x

:

\frac{dy}{dx} = 3x^2 + 4x + 1

Set

\frac{dy}{dx} = 0

:

3x^2 + 4x + 1 = 0

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

So

x = -\frac{1}{3}

or

x = -1

.

Now find the second derivative:

\frac{d^2y}{dx^2} = 6x + 4

At

x = -\frac{1}{3}

:

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

This gives a minimum.

At

x = -1

:

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

This gives a maximum.

Substitute

x = -1

into

y

:

y = -1(-1+1)^2 = 0

Answer:

The maximum value of

y

is

0

.

Correct option: B. 0

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