Question

Suppose f (x) is a differentiable function defined on the interval [0, 60]. Use the function values given in the table below to answer the following:

Answer 1

x 0 15 30 45 60
f(x) -3 -1 2 6 2

Answer 2

a) Approximate the value of f '(20).
b) Show that f '(c1) = 1/12 for some value x = c1 in the interval (0, 60).
c) Show that f (c2) = 0 for some value x = c2 in the interval (0, 60).
Justify your answers.

Answer 3

@xapproachesinfinity @zepdrix @Hero

Answer 4

@mathmale

Answer 5

not sure exactly, but for \(f'(20)\) you could take the slope between the two nearest points \((15,-1),(30,2)\)

Answer 6

That equals 5 which doesn't seem right?

Answer 7

it certainly isn't five

Answer 8

the slope between those two points i mean, definitely not 5

Answer 9

maybe \(\frac{1}{5}\)?

Answer 10

Ohhhh oops I flipped it :)

Answer 11

Do you have any idea how I would figure out b. and c.?? :)

Answer 12

my guess is a straight forward application of the mean value theorem

Answer 13

what is the slope between the two points \((0,-3)\) and \((60,2)\) ?

Answer 14

5/60=1/12?

Answer 15

lol no, not \(\frac{1}{2}\) ? but rather \[\huge \frac{1}{12}!\]

Answer 16

and that does it for you if you are familiar with the mean value theorem
did you get to that yet?

Answer 17

Yepp :)

Answer 18

What would I do for c.?? :)

Answer 19

function is continuous means it cannot skip values
it is negative somewhere, positive somewheres else, that means it has to take on every value between the positive and negative ones
that includes 0

Answer 20

How would I find where it = 0? :)

Answer 21

somewhere between where it is \(-1\) and \(2\)

Answer 22

more than this you cannot know

Answer 23

So could I just say somewhere between f(x)=-1 and 2?

Answer 24

"somewhere" means the x values
somewhere between 15 and 30

Answer 25

Oh okay, do you think that would be it for the answer then? :)

Answer 26

Thanks for all your help @satellite73 ! :)