What is the extreme sqrt(x^2+y^2-2x+26) ?

The extreme sqrt(x^2+y^2-2x+26) is Minimum(1,0)

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extreme sqrt(x^2+y^2-2x+26)

extreme points

x^{2} + y^{2} - 2 x + 26

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x ^{2} - 2 x + 26}{( x ^{2} + y ^{2} - 2 x + 26 ) x ^{2} + y ^{2} - 2 x + 26}

D (x, y) = \frac{25}{( x ^{2} + y ^{2} + 26 - 2 x ) ^{2}}

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Minimum (1, 0)

f (x) = - 1.25 x^{2} + 160 x - 2500

f (x) = 2 x^{3} - 150 x + 3

f (x, y) = x^{2} + \frac{1}{2} y^{2} + \frac{1}{2} (y - x) - \frac{3}{2}

x^{4} + 3 x^{3} + 2 x^{2} + x + 1

f (x) = - 0.4 x^{2} + 90 x - 2000, 0 \leq x

What is the extreme sqrt(x^2+y^2-2x+26) ?
The extreme sqrt(x^2+y^2-2x+26) is Minimum(1,0)