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

The extreme f(x)=(3(x^2+1))/(x^2-x+1) is Minimum(-1,2),Maximum(1,6)

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

extreme points

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

Minimum (- 1, 2), Maximum (1, 6)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, 2)

Plug

x = 1

into

\frac{3 ( x ^{2} + 1 )}{x ^{2} - x + 1} : 6

Maximum (1, 6)

Minimum (- 1, 2), Maximum (1, 6)

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

f (x) = x^{\frac{3}{2}} - x^{\frac{1}{2}}, 0 \leq x \leq 1

f (x) = - \frac{1}{4} x^{4} + x^{3} + 5

f (x) = x^{2} + 8 x + 16

f (x) = e^{2 x}

What is the extreme f(x)=(3(x^2+1))/(x^2-x+1) ?
The extreme f(x)=(3(x^2+1))/(x^2-x+1) is Minimum(-1,2),Maximum(1,6)