I was going over some practice problems and got stuck with this one: I am supposed to find the maxim

landdenaw

landdenaw

Answered question

2022-06-22

I was going over some practice problems and got stuck with this one: I am supposed to find the maximum of the function:
x x 2 + 1
on the interval (0,4).
All I can think of is to try and plug in values from the given interval . Is there a more systematic approach to finding maxima of rational functions?

Answer & Explanation

candelo6a

candelo6a

Beginner2022-06-23Added 24 answers

There is a way to do this without calculus.
You want to find the minimum value of a for which f ( x ) a for all a. In other words, the minimum value of a for which
x x 2 + 1 a
for all x [ 0 , 4 ]. Since x 2 + 1 is positive, the above inequality is equivalent to
x a x 2 + a
a x 2 x + a 0.
Call g ( x , a ) = a x 2 x + a. If g(x,a) is always positive for x in [0,4], this means that g(x,a) never crosses the x axis: it has either no real roots in g(x,a), or a double root. In both these cases, you then need to check if g(x,a) is always nonnegative (instead of always nonpositive). (You probably covered how to solve polynomial inequalities in precalculus).
The roots of g(x,a) are
x = 1 ± 1 4 a 2 2 a .
If a 1 2 , then the roots are double or not real, and g ( 0 , a ) = a 0, so all a in this region are solutions to the inequality.
If a 1 2 , then the roots also are double or not real, but this time g(0,a)<0 and the inequality is always violated.
If 1 2 < a < 1 2 , then there is a non-doubled real root in [0,4] and so the inequality does not hold for all x in [0,4].
So, finally, we've shown that f ( x ) a for all x in [0,4] precisely when a 1 2 , so the maximum of f in that region must be 1 2 . To find the value of x that gives the maximum, we plug a = 1 2 into the expression for the roots above and get
x = 1 ± 0 2 1 2 = 1.

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