What are the points of inflection of #f(x)=x^2-x^(1/2) #? Calculus Graphing with the Second Derivative Determining Points of Inflection for a Function 1 Answer Sonnhard May 26, 2018 No Points of inflection Explanation: #f(x)=x^2-x^(1/2)# #f'(x)=2x-1/2*x^(-1/2)# #f''(x)2+1/4*x^(-3/2)>0# Answer link Related questions How do you find the inflection points for the function #f(x)=8x+3-2 sin(x)#? How do you find the inflection point of a cubic function? How do you find the inflection point of a logistic function? What is the inflection point of #y=xe^x#? How do you find the inflection points for the function #f(x)=x^3+x#? How do you find the inflection points for the function #f(x)=x/(x-1)#? How do you find the inflection points for the function #f(x)=x/(x^2+9)#? How do you find the inflection points for the function #f(x)=xsqrt(5-x)#? How do you find the inflection points for the function #f(x)=e^sin(x)#? How do you find the inflection points for the function #f(x)=x-ln(x)#? See all questions in Determining Points of Inflection for a Function Impact of this question 1494 views around the world You can reuse this answer Creative Commons License