What is the extreme f(t)=(e^t)/(t^2+1),(0,3) ?

The extreme f(t)=(e^t)/(t^2+1),(0,3) is Saddle(1, e/2)

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extreme f(t)=(e^t)/(t^2+1),(0,3)

extreme points

f (t) = \frac{e ^{t}}{t ^{2} + 1}, (0, 3)

Saddle (1, \frac{e}{2})

Solution steps

Find the critical points:

t = 1

Domain of

\frac{e ^{t}}{t ^{2} + 1}, (0, 3) : 0 < t < 3

Find intervals:

Increasing

: 0 < t < 1,

Increasing

: 1 < t < 3

Plug the extreme point

x = 1

into

\frac{e ^{t}}{t ^{2} + 1} \Rightarrow y = \frac{e}{2}

Saddle (1, \frac{e}{2})

f (x) = 5 x^{3} - 5 x^{2} - 5 x + 7

x^{2} - 2 x + 1

f (x, y) = - x^{3} + 8 x y - 6 y^{2} + 12

(x - 1)^{3} + 2

f (x) = x^{5} \cdot ln (x)

What is the extreme f(t)=(e^t)/(t^2+1),(0,3) ?
The extreme f(t)=(e^t)/(t^2+1),(0,3) is Saddle(1, e/2)