Question

last question can someone please just check my work for this one problem ..

Answer 1

graph \[\sqrt{x-5}\] and see where it's differentiable

i got 5

Answer 2

yupp

Answer 3

Question is where is it differentiable? Well, it surely is differentiable for all x > 5; the question is, will it be differentiable at x = 5?

Answer: Take the derivative of sqrt(x-5), you get

y ' = 1/(2sqrt(x-5))

at if you plug in x = 5, you get an undefined expression where 0 is in the denominator, so that means at x = 5, the function is NOT differentiable.

Therefore, the function is only differentiable where x > 5, only.

Answer 4

oh jeez then i did these wrong what about (1)/(x^2 +49) is it 7 and -7?

Answer 5

or everywhere except 7 and -7

Answer 6

Do you clearly understand what I did for the last problem?

Answer 7

yes you found the derivative and plugged in 5 which gave you 1/0

Answer 8

And we also said that when x>5, the function is clearly differentiable. Do you understand why that is true?

Answer 9

because there's no value for 5 but there is a value for any number greather than 5

Answer 10

Are you 100% clear in your answer? It is critical that you understand why it is true. Otherwise, you will be stuck in further problems. Let's get this one crystal clear.

Answer 11

mmmmhmmm

Answer 12

The derivative of sqrt(x-5) is 1/(2(sqrt(x-5)).

Answer 13

So in the expression for the derivative, any number you plug in, the derivative exists! That expression has a value! That's how I know that the function is differentiable when x > 5.

Answer 14

The only point of uncertainty was at x = 5. And that I plugged into the derivative and found that the derivative is undefined at x = 5.

Answer 15

One other thing..........if you look at the graph of y = sqrt(x-5)...you will clearly see that when x > 5, the curve is smooth...so it will be differentiable there.

Answer 16

Do you clearly understand this?

Answer 17

yes :-D

Answer 18

OK...so now type any other question that you have...so we can see how that is done.

Answer 19

I think you wanted y = 1/(x^2+49)?

Answer 20

yes :-D