Thread: Help Finding if a function is differentiable or not

1)Is the function f(x) = |3x - 9| differentiable at x = 3?

2) Is the function
f(x) = {5x - 5 , x < 2
{2x + 1, x>=2
differentiable at x = 2?

for 2, when you sub in x=2, both equal 5, does that mean it is or is not differentiable?

and for 1 i'm not sure how you go about it

Originally Posted by ihatemath09

1)Is the function f(x) = |3x - 9| differentiable at x = 3?

2) Is the function
f(x) = {5x - 5 , x < 2
{2x + 1, x>=2
differentiable at x = 2?

for 2, when you sub in x=2, both equal 5, does that mean it is or is not differentiable?

and for 1 i'm not sure how you go about it

a function is differentiable at a if this limit exists:

$\displaystyle f'(a) = \displaystyle\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} $

For Q1, a = 3. Because 3 is positive, f(x) = 3x-9

For Q2, you need to evaluate the limit with both equations and they must be both exist and equal to each other to be differentiable.

Originally Posted by ihatemath09

1)Is the function f(x) = |3x - 9| differentiable at x = 3?

2) Is the function
f(x) = {5x - 5 , x < 2
{2x + 1, x>=2
differentiable at x = 2?

for 2, when you sub in x=2, both equal 5, does that mean it is or is not differentiable?

and for 1 i'm not sure how you go about it

You can test diffrentialbility by testing the limit from both sides at the point. I'd explain this in proper terms but I cant find the latex button while I'm writing this post.

Originally Posted by Gusbob

a function is differentiable at a if this limit exists:

$\displaystyle f'(a) = \displaystyle\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} $

For Q1, a = 3. Because 3 is positive, f(x) = 3x-9

No, whether a is positive or not is irrelevant. It is a question of whether 3x- 9 is positive or not. If x> 3, 3x- 9> 0 and f(x)= |3x- 9|= 3x- 9 which has slope 3. If x< 3, then 3x-9< 0 so f(x)= |3x- 9|= -(3x- 9)= -3x+ 9 which has slope -3.

ihatemath09 (if you are going to keep using this forum, you might want to change that user name!), the fact that both formulas gives the same thing at x= 2 only means the function is continuous there, not that the function is differentiable.

By the way, while the derivative of a function is not necessarily continuous, it does satisfy the "intermediate value theorem". One result of that is that if [itex]\lim_{x\to a^-} f'(x)= \lim_{x\to a^+} f'(x)[/itex] then f is differentiable at x= a. That's simpler to use but a deeper result that looking at the definition of the derivative.

What you can do is look at $\displaystyle \lim_{h\to 0^+}\frac{f(2+h)- f(2)}{h}= \lim_{h\to 0}\frac{(5(2+h)- 5)- 5}{h}$ and $\displaystyle \lim_{h\to 0^-} \frac{f(2+h)- f(2)}{h}= \lim_{h\to 0}\frac{(2(2+h)+ 1)- 5}{h}$. If those two limits are the same, then the function is differentiable at x= 2 (and the common value is the derivative there).

For Q2, you need to evaluate the limit with both equations and they must be both exist and equal to each other to be differentiable.

You can also use the alternate definition for the derivative

$\displaystyle \displaystyle \lim_{x \to a} \dfrac{f(x)-f(a)}{x-a}$

Also, I agree with HallsofIvy - I'd change a username like "Ihatemath". It might not get the help you desire.