The min value of $x^{2} - 7 x + 10 = 0$

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Solution

$x^{2} - 7 x + 10 = 0$
$f^{'} (x) = 0$ for Min value
$f^{'} (x) = 2 x - 7 = 0 x = \frac{7}{2} {(\frac{7}{2})}^{2} - 7 (\frac{7}{2}) + 10 \frac{49}{4} - \frac{49}{2} + 10 = - \frac{9}{4}$

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- x^{2} + 7 x - 10 = 0

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The min value of x2−7x+10=0

x2−7x+10=0f′(x)=0 for Min value f′(x)=2x−7=0x=72(72)2−7(72)+10494−492+10=−94

The min value of $x^{2} - 7 x + 10 = 0$

$x^{2} - 7 x + 10 = 0$
$f^{'} (x) = 0$ for Min value
$f^{'} (x) = 2 x - 7 = 0 x = \frac{7}{2} {(\frac{7}{2})}^{2} - 7 (\frac{7}{2}) + 10 \frac{49}{4} - \frac{49}{2} + 10 = - \frac{9}{4}$