Verified answer

First find the values of x for which

|f(x) - 3| = |x^2 - 4| < 0.2

Either

(a)

x^2-4 >= 0 and x^2-4 < 0.2

which implies "x <= -2 or 2 <= x" and "-sqrt(4.2) < x < sqrt(4.2)

therefore "-sqrt(4.2) < x <= -2" or "2 <= x < sqrt(4.2)"

Or

(b)

x^2-4 <= 0 and -x^2+4 < 0.2

which implies "-2 <= x <= 2" and "x < -sqrt(3.8) or sqrt(3.8) < x"

therefore "-2 <= x < -sqrt(3.8)" or "sqrt(3.8) < x <= 2"

Taking the union of these 4 regions gives

"-sqrt(4.2) < x < -sqrt(3.8)" or "sqrt(3.8) < x < sqrt(4.2)"

The condition 0 < |x-2| < delta is equivalent to

"0 < x-2 < delta" or "0 < -x+2 < delta"

which implies

"2 < x < 2+delta" or "-2 < -x < -2+delta"

which implies

"2 < x < 2+delta" or "2-delta < x < 2"

or more simply

"2-delta < x < 2+delta" (and "x is not 2")

So we need to find the largest value of delta such that the region 2-delta < x < 2+delta

lies entirely within the union of "-sqrt(4.2) < x < -sqrt(3.8)" and "sqrt(3.8) < x < sqrt(4.2)"

Clearly delta = min( 2-sqrt(3.8), sqrt(4.2)-2 ) = sqrt(4.2) - 2