First slide

Application of diffrentiation

Question

Find the minimum value of
$y = 5 x^{2} - 2 x + 1 .$

Moderate

A

$\frac{2}{5}$
B

$\frac{4}{9}$
C

$\frac{4}{5}$
D

$\frac{4}{25}$

Solution

For maximum or minimum value, $\frac{d y}{d x} = 0$
$5 (2 x) - 2 (1) + 0 = 0$
$x = \frac{1}{5}$
Now at $x = \frac{1}{5}, \frac{d^{2} y}{d x^{2}} = 10$ which is positive so minima at $x = \frac{1}{5}$ .
Therefore, $y_{\min} = 5 {(\frac{1}{5})}^{2} - 2 (\frac{1}{5}) + 1 = \frac{4}{5}$

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