What is the extreme f(x)=sqrt((x-4)^2+1)+3 ?

The extreme f(x)=sqrt((x-4)^2+1)+3 is Minimum(4,4)

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extreme f(x)=sqrt((x-4)^2+1)+3

extreme points

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

Minimum (4, 4)

Solution steps

f^{'} (x) = \frac{x - 4}{x ^{2} - 8 x + 17}

Find intervals:

Decreasing

: - \infty < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 4

into

(x - 4)^{2} + 1 + 3 : 4

Minimum (4, 4)

x^{2} y + y^{3} - 75 y

y = \frac{x ^{2}}{x ^{2} - 1}

f (x) = 6 x - 4 y - x^{2} - 2 y^{2}

f (x) = x^{4} - 6 x^{2} + 9

f (x) = - x^{2} + 8 x - 15

What is the extreme f(x)=sqrt((x-4)^2+1)+3 ?
The extreme f(x)=sqrt((x-4)^2+1)+3 is Minimum(4,4)