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DIFFERENTIATION — Statistics STD 12 Commerce — Question

Gujarat BoardEnglish MediumSTD 12 CommerceStatisticsDIFFERENTIATION3 Marks
Question
Determine whether the function $y = 2x^3 – 7x^2 – 11x + 5$ is increasing or decreasing at $x = \frac{1}{2}$ and $x = 3.$
✓

Answer

$y = 2x^3 – 7x^2 – 11x + 5$
$\therefore \frac{d y}{d x} = 2(3x^2) – 7(2x) – 11(1) + 0$
$= 6x^2 – 14x – 11$
At $x = \frac{1}{2}, \frac{d y}{d x} = 6\left(\frac{1}{2}\right)^{2} – 14\left(\frac{1}{2}\right) – 11$
$= \frac{6}{4} – 7 – 11$
$= \frac{6}{4} – 18$
$= \frac{6 -72}{4}$
$= –\frac{66}{4}, < 0$
So, at $x = \frac{1}{2}$ function is decreasing.
At $x = 3, \frac{d y}{d x} = 6(3)^2 – 14(3) – 11$
$= (6 \times 9) – 42 – 11$
$= 54 – 53 = 1 > 0$
So, at $x = 3$ function is increasing.

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