Question

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

Answer 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!

Answer 2

I would apologize for being pedantic, but this is an educational website.

Explanation:

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]}