What are the local extrema, if any, of #f(x) =2-x^2#?

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Answer 1 · 2016-10-01T02:31:09

#f(x)# has a maximum value of #2# at #x=0#

#f(x) = 2-x^2#

#f'(x) = -2x#

For a local maximum or minimum #f'(x) = 0#

# -2x = 0 -> x=0#

Hence #f(0) =2# is a local maximum or minimum.

Now consider #f''(x) =-2# #<0# independent of #x#

So, #f(0) = 2# is a maximum value of #f(x)#

This can be seen on the graph of #2-x^2# below:

graph{2-x^2 [-4.933, 4.934, -2.467, 2.465]}