How do you sketch the graph that satisfies f'(x)>0 when x does not equal 2, f(2)=1?

Since the derivative is greater than #0# on all #x# excluding #x = 2#, we know the function is increasing until it gets to #x= 2#, where it plateaus, and then it starts increasing again.

A perfect example of this would be the cubic function #f(x) = (x - 2)^3 + 1#, as pictured in the following graph.

Hopefully this helps!

The question does not give any information about #f'(2)#.

It is important to understand the use of language and logic in mathematics.
Saying #f'(x) > 0# when #x != 2# DOES NOT entail that #f'(2) = 0# or even that #f'(2)# exists.

#f'(2)# cannot be negative (because a function that is differentiable on an interval satisfies the intermediate value property), but it could be #0# or positive or it could fail to exist.

So,

(1) Any line with positive slope through #(2,1)# meets the condition

for example #f(x)=3(x-2)+1# has #f'(x) = 3 > 0# for all #x != 2# and also for #x = 2#

graph{y-1=3(x-2) [-1.907, 9.19, -2.45, 3.1]}

Replace #3# with any positive number for another example.

Any other curve with positive slope everywhere will also work, for example #f(x) - e^(x-2)# has positive slope everywhere and contains the point #(2,1)#.

graph{e^(x-2) [-1.29, 8.574, -2.14, 2.793]}

(2) We could also have a piecewise function with positive slope except at a discontinuity at #x=2#.

#f(x) = {(5x+4,x != 2),(1,x = 2):}#

This has #f'(x) = 5# for all #x != 2# and #f'(2) does not exist.

Or

#f(x) = (x-2)^(1/3)+1# which has #f'(x) = 1/2(x-2)^(-2/3) > 0# for #x !=2# and #f'(2)# does not exists (vertical tangent line).

graph{(x-2)^(1/3)+1 [-2.68, 5.113, -0.86, 3.037]}

(3) Or we could have a translation of an odd power function.

#f(x) = (x-2)^3+1# or #f(x) = (x-2)^5+1# etc.

These have #f'(x) >0 # for #x !=2# and #f'(2) =0#

graph{(x-2)^(7/3)+1 [-1.216, 3.65, -0.128, 2.305]}