APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES2 Marks
Question
Find the maximum and minimum values, if any, of the function given by:
$f(x) = 9x^2 + 12x + 2$
✓
Answer
Given: $\text{f}\text{(x)}=9\text{x}^2+12+2$
$\Rightarrow\ \text{f}\text{(x)}=9\Big(\text{x}^2+\frac{4\text{x}}{3}+\frac{2}{9}\Big)$
$\Rightarrow \ \text{f}\text{(x})=9\Bigg(\text{x}^2+\frac{4\text{x}}{3}+\Big(\frac{2}{3}\Big)^2-\Big(\frac{2}{3}\Big)^2+\frac{2}{9}\Bigg)$
$=9\Bigg[\Big(\text{x}+\frac{2}{3}\Big)^2-\frac{4}{9}+\frac{2}{9}\Bigg]$
$\Rightarrow \ \text{f}\text{(x)}=9\Big(\text{x}+\frac{2}{3}\Big)^2-2\ \dots\text{(i)}$
$\text{Since }\ 9\Big(\text{x}+\frac{2}{3}\Big)^2 \geq0\text{ for all } \text{x}\in \text{R}$
Subtracting 2 from both sides, $9\Big(\text{x}+\frac{2}{3}\Big) -2\geq0-2\ \Rightarrow\ \text{f}\text{(x)}\geq-2$
Therefore, minimum value of f(x) is -2 and is obtained when $\text{x}+\frac{2}{3}=0, \text{i. r.,}\text{ x}= \frac{-2}{3}$
And this function does not have a maximum value.
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