What are the absolute extrema of # f(x)= 2 + x^2 in [-2, 3]#? Calculus Graphing with the First Derivative Identifying Turning Points (Local Extrema) for a Function 1 Answer Alan N. · Anjali G Mar 5, 2017 Answer: #f(x)# has an absolute minimum of 2 at #x=0# Explanation: #f(x)= 2+x^2# #f(x)# is a parabola with a single absolute minimum where #f'(x)=0# #f'(x) =0+2x = 0 -> x=0# #:.f_min(x) = f(0) = 2# This can be seen on the graph of #f(x)# below: graph{2+x^2 [-9.19, 8.59, -0.97, 7.926]} Related questions How do you find the x coordinates of the turning points of the function? How do you find the turning points of a cubic function? How many turning points can a cubic function have? How do you find the coordinates of the local extrema of the function? How do you find the local extrema of a function? How many local extrema can a cubic function have? How do I find the maximum and minimum values of the function #f(x) = x - 2 sin (x)# on the interval ... If #f(x)=(x^2+36)/(2x), 1 <=x<=12#, at what point is f(x) at a minimum? How do you find the maximum of #f(x) = 2sin(x^2)#? How do you find a local minimum of a graph using the first derivative? See all questions in Identifying Turning Points (Local Extrema) for a Function Impact of this question 239 views around the world You can reuse this answer Creative Commons License