# Extreme Values on non-closed intervals?

a.) f(x)= (x^2 +1) / x on (0,2]

b.) f(x)= x / (x^2 + 4) on [-4, infinity)

If anyone can go through these an explain how to answer the problem that would be very helpful!

^^I need help with how to find the extreme values for the equations

### 1 Answer

- AlanLv 72 months agoFavourite answer
For extremes on an interval , you have to check

the endpoints values (even if it doesn't reach the endpoint,

because they may be no minimium or maximium)

a)

f(0) = undefined but it approaches +infinitiy

so the extreme value will infinity or no maximum

so infinity can never be reached.

f(2) = (4+1)/2 = 5/2 = 2.5

so then , check for critical points and/or discontinuities

d( x^2 + 1)/ x) = (x* 2x - (x^2+1) ) / x^2 = (x^2-1)/x^2

so this equals 0

when x^2 = 1

x =+/-1

only x=+1 is in our range

f(1) = (1+1) /1 = 2

so use 2nd

derivative test

f''(x) = (x^2 *2x - (x^2-1)*2x) / x^4 = (2x^3 - 2x^3 +2x)/x^4

f"(x) = +2x/x^4

f(1) = 2*1/1 = 2 (positive it is a minimum

the maximum value approaches +infinity

Since you can never reach infinity

most math textbook so they is no maximum

the minimum is at ( 1,2)

(b)

f(x)= x / (x^2 + 4) [-4,infinity)

check endpoints again

f(-4) = -4/ ( 16+4) = -4/20 = -1/5

f(+infinity) = x/x^2 =1/x so it approaches 0

f'(x) = ( (x^2+4)*1 - 2x*x ) / (x^2+4)

f'(x) = ( 4-x^2 ) / (x^2 +4)

f'(x) = 0

when

4 -x^2 = 0

x^2 = 4

x = +/- 2

f(2) = 2/ (4+4) = 1/4

f(-2) = -2/ (4 + 4) = -1/4

so -1/4< -1/5 < 0 < 1/4

so -1/4 is the minimum value

+1/4 is the maximum values

minimum (-2, -1/4)

maximum (2, 1/4)

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