Given, 3( x−2 ) 5 ≤ 5( 2−x ) 3 .
Solve for x,
3( x−2 ) 5 ≤ 5( 2−x ) 3 3x−6 5 ≤ 10−5x 3 3( 3x−6 )≤5( 10−5x ) 9x−18≤50−25x
Further simplify,
9x+25x≤50+18 34x≤68 x≤ 68 34 x≤2
All the real numbers less than or equal to 2 are the solutions of inequality, as x is a real number.
Therefore, solution for inequality is ( −∞,2 ].
The maximum and minimum values of f (x) = x50-x20 in the interval [0, 1] are