Question

Given f(x)= {5x-6, x<2
{x^2, 24
find f'(x) and give its domain

My question is how does taking the derivative affect the restrictions?

Answer 1

@pooja195

Answer 2

@zepdrix @jagr2713

Answer 3

@freckles @dan815

Answer 4

the endpoints need to be check to see if left derivative=right derivative

Answer 5

for example for f' to exist at x=2 you must have (5x-6)'=(x^2)' as x goes to 2

Answer 6

you will need to do something similar for x=4

Answer 7

oh you closed this
does that mean you figured it out?

Answer 8

no i didn't figure it out

Answer 9

ill try to take a screenshot of the problem

Answer 10

are you understanding what I'm saying though?

Answer 11

f' is definitely going to exist for all x
except for x=2 or x=4 because all the functions are continuous for all real numbers
for x=2 or x=4 we are not sure in this case
we have to check to see if the left derivative=right derivative for both x=2 and x=4

Answer 12

so its a limit problem? i need to look at whether theres a limit at 2 and 4?

Answer 13

\[\lim_{x \rightarrow 2}(5x-6)'=\lim_{x \rightarrow 2}(x^2)' \text{ if this holds then yes } \\ f' \text{ exists at } x=2 \]

\[\lim_{x \rightarrow 4}(x^2)'=\lim_{x \rightarrow 4}(2x+8)' \text{ if this holds then } f' \text{ exists at } x=4\]

you are checking to see if the derivative is continuous at these numbers
we are doing exactly what would do to see if the original function was continuous at x=2 or x=4 which it is since 5*2-6=2^2 and 4^2=2*4+8
which should have been the first thing we looked at but anyways back to the above limits

Answer 14

so the first one does exist because its 4=4
and the second one exists too because its 16=16?

Answer 15

so the answer is B, domain= all real numbers?

Answer 16

how did you get that?

Answer 17

i plugged in for x for each equation

Answer 18

after finding the derivative right?

Answer 19

I already checked above to see if the original function was continuous at x=2 or x=4

again we need to check to see if the derivative function is continuous at x=2 or x=4

Answer 20

the derivative of the first would equal 5 and of the second 2x?

Answer 21

\[\lim_{x \rightarrow 2}(5x-6)'=\lim_{x \rightarrow 2}(x^2)' \\ \lim_{x \rightarrow 2}(5)=\lim_{x \rightarrow 2}(2x)\]
ok now see if this equality holds

Answer 22

no because it would be 5=4

Answer 23

right so f' will definitely not exist at x=2 since the slopes do not match up at x=2

Answer 24

so it would exist at 4 either because the derivative of the first is 2x and 2 for the second. so it would be 8=2

Answer 25

so the answer is A

Answer 26

right

To find if f'(a) exists:
see if f is continuous at a
and if yes continue to see if f' is continuous at a

Answer 27

ok can you help with one more please?

Answer 28

might be able to

Answer 29

and you graphed it?

Answer 30

yes, im not sure what it means increases without bound

Answer 31

it looks like this

Answer 32

it means it function increases and never stops

Answer 33

but wouldnt there also be a tangent line of 0 at the maximum?

Answer 34

what kind of tangent line ? :)

Answer 35

wouldn't that be a tangent line? the tangent line is the slope of a derivative?