If u (x) = negative 2 x squared and v (x) = StartFraction 1 Over x EndFraction, what is the range of (u circle v) (x)?
2 answers:
Answer:
(-∞, 0)
Step-by-step explanation:
<u>Given</u>
u(x) = -2x^2
v(x) = 1/x
<u>Find</u>
The range of (u◦v)(x)
<u>Solution</u>
(u◦v)(x) = u(v(x)) = u(1/x) = -2(-1/x)^2 = -2/x^2
The denominator is always positive (since x=0 is excluded from the domain), so the fraction is always negative. There is a horizontal asymptote at y=0. The range is ...
(-∞, 0)
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Given
Your family has a circular swimming pool.
The radius of the pool is 9 feet and and its depth is 4 feet.
Find out the volume of the pool.
To proof
As depth is given in the question that means the shape of the swimming pool
is cylinder.
Formula
where r = radius
h = height
now
radius of the pool is 9 feet and and its depth is 4 feet.
put all the values in the above equation
we get
use 3.14 to approximate pi
we get
= 1017.4 ( approx )feet²
Hence proved