Question

x is real then the minimum value of x^2-6x+10

Answer 1

Answer:

Min value is 1.

Step-by-step explanation:

Let Z = x² - 6 x + 10 .

Differentiate Z with respect to variable x.

dZ/dx = 2 x - 6

Equate it to 0 to get the value of x at which point Z is minimum or maximum.

So 2 x - 6 = 0

x = 3.

Z at x = 3 is 3² - 6 * 3 + 10 = 1.

Find second derivative of Z wrt x.

d²Z/dx² = 2   > 0  

So Z is minimum at x = 3.

Answer 2

y=x²-6x+10

$\frac{dy}{dx} = 2x - 6 \\ \\ we \: will \: put \: value \: \frac{dy}{dx} = 0 \: and \: find \: to \: value \: of \: x \\ 2x - 6 = 0 \\ \\ 2x = 6 \\ \\ x = 3 \\ \\ \\ x = 3 \: value \: put \: in \: main \: equation \\ \\ \\ y(3) = {3}^{2} - 6(3) + 10 \\ \\ \\ \: \: \: \: \: \: \: \: \: = 9 - 18 + 10 \\ \\ \\ so \: your \: answer \: is \: 1$