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

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

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extreme f(t)=(t^2-t+1)/(t^2+1)

extreme points

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

Maximum (- 1, \frac{3}{2}), Minimum (1, \frac{1}{2})

Solution steps

f^{'} (t) = \frac{t ^{2} - 1}{( t ^{2} + 1 ) ^{2}}

Find intervals:

Increasing

: - \infty < t < - 1,

Decreasing

: - 1 < t < 1,

Increasing

: 1 < t < \infty

Plug

t = - 1

into

\frac{t ^{2} - t + 1}{t ^{2} + 1} : \frac{3}{2}

Maximum (- 1, \frac{3}{2})

Plug

t = 1

into

\frac{t ^{2} - t + 1}{t ^{2} + 1} : \frac{1}{2}

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

Maximum (- 1, \frac{3}{2}), Minimum (1, \frac{1}{2})

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

x e^{- 6 x^{2}}

f (x) = x^{4} - 8 x^{3} + 10 x^{2} + 1

- 15 x^{4} + 120 x^{2}

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

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