What is the 36y^2=(x^2-4)^3,5<= x<= 9 ?

The 36y^2=(x^2-4)^3,5<= x<= 9 is

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36y^2=(x^2-4)^3,5<= x<= 9

36 y^{2} = (x^{2} - 4)^{3}, 5 \leq x \leq 9

Domain : 5 \leq x \leq 9

Range : 9261 \leq f (x) \leq 456533

Intercepts : None

Asymptotes : None

Extreme Points : Minimum (5, 9261), Maximum (9, 456533)

Interval Notation

Domain : [5, 9]

Range : [9261, 456533]

Solution steps

Domain of

(x^{2} - 4)^{3}, 5 \leq x \leq 9 : 5 \leq x \leq 9

Range of

(x^{2} - 4)^{3}, 5 \leq x \leq 9 : 9261 \leq f (x) \leq 456533

Axis interception points of

(x^{2} - 4)^{3}, 5 \leq x \leq 9 :

None

Asymptotes of

(x^{2} - 4)^{3}, 5 \leq x \leq 9 :

None

Extreme Points of

(x^{2} - 4)^{3}, 5 \leq x \leq 9 :

Minimum

(5, 9261),

Maximum

(9, 456533)

f (x) = \frac{x}{a ^{2} - x ^{2}}

f (t) = 5 t^{7} + cos (3 t) - e^{- 2 t}

f (x) = e^{\frac{3 x ^{2}}{2}}

y = \frac{v ^{3} - 2 v v}{v}

f (x) = a x + \frac{a}{a x}