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Anonymous
Anonymous asked in Education & ReferenceHomework Help · 2 months ago

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!

Update:

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

1 Answer

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  • Alan
    Lv 7
    2 months ago
    Favourite 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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