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What are the maximum and minimum values that the function #f(x)=x/(1 + x^2)#?

Answer 1

William Abdalla

Maximum: #1/2#
Minimum: #-1/2#

An alternative approach is to rearrange the function into a quadratic equation. Like this:

#f(x)=x/(1+x^2)rarrf(x)x^2+f(x)=xrarrf(x)x^2-x+f(x)=0#

Let #f(x)=c" "# to make it look neater :-)

#=> cx^2-x+c=0#

Recall that for all real roots of this equation the discriminant is positive or zero

So we have, #(-1)^2-4(c)(c)>=0" "=>4c^2-1<=0" "=>(2c-1)(2c+1)<=0#

It is easy to recognise that #-1/2<=c<=1/2#

Hence, #-1/2<=f(x)<=1/2#

This shows that the maximum is #f(x)= 1/2# and the minimum is #f(x)=1/2#

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Answer 2

Ezra Adams

The maximum value of the function f(x) = x/(1 + x^2) occurs as x approaches positive or negative infinity and equals 1. The minimum value of the function does not exist because as x approaches positive or negative infinity, f(x) approaches 0 but does not reach a definite minimum value.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.