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.