What is the extreme |x^2-5|,1<x<7 ?

The extreme |x^2-5|,1<x<7 is Minimum(sqrt(5),0)

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extreme |x^2-5|,1<x<7

extreme points

x^{2} - 5, 1 < x < 7

Minimum (5, 0)

Solution steps

Find the critical points:

x = 5

Find intervals:

Decreasing

: 1 < x < 5,

Increasing

: 5 < x < 7

Plug the extreme point

x = 5

into

x^{2} - 5 \Rightarrow y = 0

Minimum (5, 0)

(x - 7) (x - 5)^{2} (x - 3) (x - 1)

\frac{400}{1 + 8 e ^{0.05 x}}

x^{4} - 50 x^{2} + 2

1 + (ln (x))

f (x, y) = 3 x^{2} + 3 x - 2 y + 4 y^{2} - 12 x y

What is the extreme |x^2-5|,1<x<7 ?
The extreme |x^2-5|,1<x<7 is Minimum(sqrt(5),0)