What is the V(r)=pir^2((500-pir^2)/(2pir)) ?

The V(r)=pir^2((500-pir^2)/(2pir)) is

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V(r)=pir^2((500-pir^2)/(2pir))

V (r) = π r^{2} (\frac{500 - π r ^{2}}{2 π r})

Domain : r < 0 or r > 0

Range : - \infty < f (r) < \infty

X Intercepts : (- \frac{10 5}{π}, 0), (\frac{10 5}{π}, 0)

Asymptotes : None

Extreme Points : Minimum (- \frac{10 5}{3 π}, - \frac{5000 5}{3 3 π}), Maximum (\frac{10 5}{3 π}, \frac{5000 5}{3 3 π})

Interval Notation

Domain : (- \infty, 0) \cup (0, \infty)

Range : (- \infty, \infty)

Solution steps

Domain of

π r^{2} \frac{500 - π r ^{2}}{2 π r} : r < 0 or r > 0

Range of

π r^{2} \frac{500 - π r ^{2}}{2 π r} : - \infty < f (r) < \infty

Axis interception points of

π r^{2} \frac{500 - π r ^{2}}{2 π r} :

X Intercepts

: (- \frac{10 5}{π}, 0), (\frac{10 5}{π}, 0)

Asymptotes of

π r^{2} \frac{500 - π r ^{2}}{2 π r} :

None

Extreme Points of

π r^{2} \frac{500 - π r ^{2}}{2 π r} :

Minimum

(- \frac{10 5}{3 π}, - \frac{5000 5}{3 3 π}),

Maximum

(\frac{10 5}{3 π}, \frac{5000 5}{3 3 π})

f (x) = x^{2} - 5 x

g (x) = 5^{- x}

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

y = (x - 4)^{2} - 5

f (x) = \frac{6}{x}

What is the V(r)=pir^2((500-pir^2)/(2pir)) ?
The V(r)=pir^2((500-pir^2)/(2pir)) is