What is the extreme f(x)=x^2-x+5 ?

The extreme f(x)=x^2-x+5 is Minimum(1/2 , 19/4)

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Find the Local Maxima and Minima f(x)=x^2-x+5

extreme points

f (x) = x^{2} - x + 5

Minimum (\frac{1}{2}, \frac{19}{4})

Solution steps

f^{'} (x) = 2 x - 1

Find intervals:

Decreasing

: - \infty < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = \frac{1}{2}

into

x^{2} - x + 5 : \frac{19}{4}

Minimum (\frac{1}{2}, \frac{19}{4})

y = arctan (x)

- 3 y = - 12

3 x - 4 y = - 8

y - 4 x = 7 - x^{2}

f (x) = x^{3} + 7

What is the extreme f(x)=x^2-x+5 ?
The extreme f(x)=x^2-x+5 is Minimum(1/2 , 19/4)