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Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = (36 − t^2)^ 1/t, [−1, 6]Asked by TayB
Find the absolute maximum and absolute minimum values of f on the given interval.
f(x)=(x^2-4)/(x^2+4), [-4,4]
f(x)=(x^2-4)/(x^2+4), [-4,4]
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Answered by
Reiny
use the quotient rule,
f ' (x) = ( 2x(x^2 + 4) - 2x(x^2 - 4))/(x^2 + 4)^2
= (2x^3 + 8x - 2x^3 + 8x)/(x^2 + 4)^2
= 16x/(x^2+4)^2
= 0
---> x = 0
if x = 0 , f(0) = -4/4 = -1
(0,-1) is a local min by looking at the graph
http://www.wolframalpha.com/input/?i=f%28x%29%3D%28x%5E2-4%29%2F%28x%5E2%2B4%29
but since there is a domain, we have to look at the end points of the domain
by summetry,
f(±4) = (16-4)/(16+4)
= 12/20
= 3/5
so (-4,3/5) and (4,3/5) are maximum points
since it asked for the min and max value of f
the min is -1, and the max is 3/5
f ' (x) = ( 2x(x^2 + 4) - 2x(x^2 - 4))/(x^2 + 4)^2
= (2x^3 + 8x - 2x^3 + 8x)/(x^2 + 4)^2
= 16x/(x^2+4)^2
= 0
---> x = 0
if x = 0 , f(0) = -4/4 = -1
(0,-1) is a local min by looking at the graph
http://www.wolframalpha.com/input/?i=f%28x%29%3D%28x%5E2-4%29%2F%28x%5E2%2B4%29
but since there is a domain, we have to look at the end points of the domain
by summetry,
f(±4) = (16-4)/(16+4)
= 12/20
= 3/5
so (-4,3/5) and (4,3/5) are maximum points
since it asked for the min and max value of f
the min is -1, and the max is 3/5
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