What is the f(t)=sqrt(t^3)-(sqrt(t^9))/4+e ?

The f(t)=sqrt(t^3)-(sqrt(t^9))/4+e is

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f(t)=sqrt(t^3)-(sqrt(t^9))/4+e

f (t) = t^{3} - \frac{t ^{9}}{4} + e

Domain : t \geq 0

Asymptotes : None

Solution steps

Domain of

t^{3} - \frac{t ^{9}}{4} + e : t \geq 0

Axis interception points of

t^{3} - \frac{t ^{9}}{4} + e :

X Intercepts

Y Intercepts

: (0, e)

Asymptotes of

t^{3} - \frac{t ^{9}}{4} + e :

None

Extreme Points of

t^{3} - \frac{t ^{9}}{4} + e :

Minimum

(0, e),

Maximum

f (x) = 0.5 x^{2} - 10 x + 70

y = - (x - 3)^{5} (x + 1)^{4}

f (x) = - - x

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

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

What is the f(t)=sqrt(t^3)-(sqrt(t^9))/4+e ?
The f(t)=sqrt(t^3)-(sqrt(t^9))/4+e is