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

2 Answers
Dec 20, 2017

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.

enter image source here

Hopefully this helps!

Dec 20, 2017

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]}