Find the value of x at the minimum point of the... | JAMB UTME - Mathematics - 2011

Question 1 Report

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

\frac{1}{3}

\frac{1}{3}

-1

Answer Details

y = x3 + x2 - x + 1
$\frac{d y}{d x}$ = $\frac{d (x^{3})}{d x}$ + $\frac{d (x^{2})}{d x}$ - $\frac{d (x)}{d x}$ + $\frac{d (1)}{d x}$
$\frac{d y}{d x}$ = 3x2 + 2x - 1 = 0
$\frac{d y}{d x}$ = 3x2 + 2x - 1
At the maximum point $\frac{d y}{d x}$ = 0
3x2 + 2x - 1 = 0
(3x2 + 3x) - (x - 1) = 0
3x(x + 1) -1(x + 1) = 0
(3x - 1)(x + 1) = 0
therefore x = $\frac{1}{3}$ or -1
For the maximum point
$\frac{d^{2} y}{d x^{2}}$ < 0

$\frac{d^{2} y}{d x^{2}}$ 6x + 2
when x = $\frac{1}{3}$
$\frac{d x^{2}}{d x^{2}}$ = 6( $\frac{1}{3}$ ) + 2
= 2 + 2 = 4
$\frac{d^{2} y}{d x^{2}}$ > o which is the minimum point
when x = -1
$\frac{d^{2} y}{d x^{2}}$ = 6(-1) + 2
= -6 + 2 = -4
-4 < 0

therefore, $\frac{d^{2} y}{d x^{2}}$ < 0

the maximum point is -1

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Differentiation

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

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