What is the extreme f(x)=sqrt(2x^2-4x+7) ?

The extreme f(x)=sqrt(2x^2-4x+7) is Minimum(1,sqrt(5))

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extreme f(x)=sqrt(2x^2-4x+7)

extreme points

f (x) = 2 x^{2} - 4 x + 7

Minimum (1, 5)

Solution steps

f^{'} (x) = \frac{2 ( x - 1 )}{2 x ^{2} - 4 x + 7}

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 1

into

2 x^{2} - 4 x + 7 : 5

Minimum (1, 5)

f (x) = \frac{( ( x + 4 ) ( x - 3 ) ^{2} )}{x ^{4} ( x - 1 )}

f (x) = 615 t - \frac{15}{5} t^{2}

f (x) = \frac{2 x ^{2} + 1}{x ^{2} - 4}

f (t) = t \cdot e^{- ∣ x ∣}

f (x) = \frac{x ^{4}}{4} + x^{3} - 14 x^{2} - 60 x

What is the extreme f(x)=sqrt(2x^2-4x+7) ?
The extreme f(x)=sqrt(2x^2-4x+7) is Minimum(1,sqrt(5))