# Multiple ChoiceMultiple Choice: Given that ` f(x)=|x-3|+2` , which one of the following statements is false?  Support your answer. (A)  f is continuous at x=3 (B)  f is differentiable at x=3...

Multiple Choice

Multiple Choice:

Given that ` f(x)=|x-3|+2` , which one of the following statements is false?  Support your answer.

(A)  f is continuous at x=3

(B)  f is differentiable at x=3

(C) `f'(5)=1`

(D)` f'(0)=-1`

(E) `f(2)=f(4)`

txmedteach | Certified Educator

Let's go through each question option one by one.

(A) f is continuous at x = 3

In order for f to be continuous, the following must hold true:

`lim_(x->3) f(x) = f(3)`

In order for the above limit to exist, the limits from the left and right must exist, as well:

`lim_(x->3^+) f(x) = lim_(x->3^-)f(x)`

Let's start by calculating the limit on the left:

`lim_(x->3^+) |x-3|+2`

If we are approaching from the left, then `x-3` will be negative. This is easily verified by checking any `x<3` and finding the result of `x-3`. Because the term inside the absolute value will be negative, we must take the negative of what is inside the absolute value to find a positive result:

`lim_(x->3+)(-x+3)+2 = lim_(x->3+) 5-x = 2`

So, our limit from the left will be 2. Now to calculate our limit from the right:

`lim_(x->3^-) |x-3|+2`

From the right, the term in the absolute value is positive, so we can simply drop the absolute value bars!

`lim_(x->3^-) x-3+2 = lim_(x->3^-) x-1 = 2`

Therefore, the limit from the right is also two. These two give us the following result:

`lim_(x->3) |x-3|+2 = 2`

Finally, we need to calculate f(3):

`f(3) = |3-3|+2 = 0+2 = 2`

Therefore, we find the desired result:

`lim_(x->3)f(x) = f(3)`

Therefore, the curve is continuous at x=3, and statement (A) is true.

(B) f is differentiable at 3

In order for f to be differentiable at `x=3`, the limit of the derivative from the left and right must be continuous:

`lim_(x->3^+)f'(x) = lim_(x->3^-)f'(x)`

Well, we already have our functions from the left and the right as they approch 3 (see above where we were calculating their limits). Now, we just need to put those same functions into these limits and take their derivative with respect to x:

`lim_(x->3^+) d/dx (5-x) = lim_(x->3^-) d/dx (x-1)`

Calculating the derivatives:

`lim_(x->3^+) -1 = lim_(x->3^-) 1`

This statement is equivalent to saying 1 = -1. Therefore, the function is not differentiable at `x = 3`. Therefore, (B) is false.

The rest of the answers can be shown to be true through rote calculation. Recognize that the derivative of the function will be of the following form:

`f'(x) =sgn(x) = x/|x| = {(-1 if x<3),(1 if x>3):}`

These different notations are all the same thing, so don't worry about never seeing sgn(x) before! This can be confirmed by noticing the slopes of the lines of the absolute value function before and after x = 3.

Another way to think about this problem is that it is a favorite of teachers, professors, and AP tests to check whether you understand that the maximal or minimal point of an absolute value function has no derivative. Just something to watch out for!

Good luck!