We know from part (c) that f ″ ( x ) {\displaystyle f''(x)} changes sign at x = 5 / 2 {\displaystyle x=5/2} , and we have f ( 5 / 2 ) = ( 5 / 2 ) 9 / 7 + ( 9 / 2 ) ( 5 / 2 ) 2 / 7 = 7 ⋅ ( 5 / 2 ) 2 / 7 {\displaystyle f(5/2)=(5/2)^{9/7}+(9/2)(5/2)^{2/7}=7\cdot (5/2)^{2/7}} . Therefore, the only inflection point is at ( x , y ) = ( 5 / 2 , 7 ⋅ ( 5 / 2 ) 2 / 7 ) {\displaystyle \color {blue}(x,y)=\left(5/2,7\cdot (5/2)^{2/7}\right)} .
