How do I find all the critical points of #f(x)=(x-1)^2#?

1 Answer
Jim H
Oct 29, 2015

See the explanation.

Explanation:

The critical numbers for a function #f# and the numbers in the domain of #f#, at which either #f(x)# does not exist or #f(x)=0#.

For #f(x)=(x-1)^2#, note that the domain is #(-oo,oo)#.

Find #f'(x)#
We can use the power and chain rule to get:

#f'(x) = 2(x-1)^1 d/dx(x-1) = 2(x-1)(1) = 2(x-1)#

Or we can expand #f(x) = x^2-2x+1# and differentiate this polynomial:

#f'(x) = 2x-2#
Find the critical numbers for #f#

Either way, we see that
#f'(x) is never undefined and #f'(x) = 0# at #x=1#.

#1# is in the domain of #f#, so it is a critical numbers for #f#.

The "critical points" are (most commonly), the same as critical numbers.
But some use a variant terminology in which critical points are points on the graph, so the have both #x# and #y# coordinates.

If you are in such a situation, you need to find #y# when #x=1# to finish.

#y=f(1) = 0#, so the point on the graph is #(1,0)#

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