Find the local and absolute extreme values of the function on the given interval.  f(x)=3x/2x^2+5,  [0,3]
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Find the local and absolute extreme values of the function on the given interval.  f(x)=3x/2x^2+5,  [0,3]

The funcrtion is: .  f(x)=3x/(2x^2+5),  on the interval [0,3]

At x = 0, f = 0. At x = 3, f = 9/(18+5) = 9/23 = 0.3913

The extreme values, as ordered pairs, are: (0, 0), (3, 9/23)

For the local extremum, get df/dx and set it to zero.

df/dx = 3/(2x^2 + 5) – (4x.3x)/(2x^2 + 5)^2

df/dx = {3.(2x^2 + 5) – (12x^2)}/(2x^2 + 5)^2

df/dx = {6x^2 + 15 – 12x^2}/(2x^2 + 5)^2

df/dx = {15 – 6x^2}/(2x^2 + 5)^2 = 0

i.e. 15 – 6x^2 = 0

x^2 = 5/2

x = √(2.5) = 1.5811

At x = √(2.5), f(x) = 3√(2.5)/(2*2.5 + 5) = 3√(2.5)/10 = 0.4743

The value of the local extremun, at x = 1.5811, is f = 0.4743

Since the local extremum is greater than the two extreme values then the extremum is a local maximum.

by Level 11 User (81.5k points)

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