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

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 $f \left(x\right) = {\left(x - 2\right)}^{3} + 1$, as pictured in the following graph.

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 $f ' \left(2\right)$.

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

$f ' \left(2\right)$ 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 $\left(2 , 1\right)$ meets the condition

for example $f \left(x\right) = 3 \left(x - 2\right) + 1$ has $f ' \left(x\right) = 3 > 0$ for all $x \ne 2$ 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 $f \left(x\right) - {e}^{x - 2}$ has positive slope everywhere and contains the point $\left(2 , 1\right)$.

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 $x = 2$.

$f \left(x\right) = \left\{\begin{matrix}5 x + 4 & x \ne 2 \\ 1 & x = 2\end{matrix}\right.$

This has $f ' \left(x\right) = 5$ for all $x \ne 2$ and #f'(2) does not exist.

Or

$f \left(x\right) = {\left(x - 2\right)}^{\frac{1}{3}} + 1$ which has $f ' \left(x\right) = \frac{1}{2} {\left(x - 2\right)}^{- \frac{2}{3}} > 0$ for $x \ne 2$ and $f ' \left(2\right)$ 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.

$f \left(x\right) = {\left(x - 2\right)}^{3} + 1$ or $f \left(x\right) = {\left(x - 2\right)}^{5} + 1$ etc.

These have $f ' \left(x\right) > 0$ for $x \ne 2$ and $f ' \left(2\right) = 0$

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