# Answer to Question #17669 in Calculus for hsd

Question #17669

Find the absolute and local maximum and minimum values of f(x)={x^2 −1≤x<0

{2−x^2 0≤x≤1.

If there are multiple points in a single category list the points in increasing order in x value and enter N in any blank that you don't need to use.

Absolute maxima

x = y =

x = y =

x = y =

Absolute minima

x = y =

x = y =

x = y =

Local maxima

x = y =

x = y =

x = y =

Local minima

x = y =

x = y =

x = y =

{2−x^2 0≤x≤1.

If there are multiple points in a single category list the points in increasing order in x value and enter N in any blank that you don't need to use.

Absolute maxima

x = y =

x = y =

x = y =

Absolute minima

x = y =

x = y =

x = y =

Local maxima

x = y =

x = y =

x = y =

Local minima

x = y =

x = y =

x = y =

Expert's answer

we must find f'

f'=2x when -1<x<0

f'=-2x when 0<x<1

f'=0 x=0

to find absolute maxima and minima we must find values atthe ends, at points of discontinuity and at points where f'=0

x=0 f=2

x=-1 f=0

x=1 f=1

and when x<0 and x->0 we have that f->-1 soabsolute maxima x=0 y=f(x)=2 absolute minima does not exist x->0 we have that f->-1 local minimax=-1 f=0, x=1 f=1 local maxima x=0 y=f(x)=2

f'=2x when -1<x<0

f'=-2x when 0<x<1

f'=0 x=0

to find absolute maxima and minima we must find values atthe ends, at points of discontinuity and at points where f'=0

x=0 f=2

x=-1 f=0

x=1 f=1

and when x<0 and x->0 we have that f->-1 soabsolute maxima x=0 y=f(x)=2 absolute minima does not exist x->0 we have that f->-1 local minimax=-1 f=0, x=1 f=1 local maxima x=0 y=f(x)=2

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