Question

I am in Stewarts Calc and am stuck on the on a derative problem...
It reads:
Let f(x)=x^2
(a) Estimate the values of f'(0), f'(1/2), f'(1), f'(2) by using a graphing device to zoom in on the graph of "f".

Answer 1

id rather just derive x^2 and input the given values

Answer 2

but that is not how the teachers book
dose it....

Answer 3

What's the problem? You put y=x^2 into your graphic calculator, look at the graph and estimate the slopes?

Answer 4

you have to learn to stand on your one 2 synapses :)

Answer 5

they graph it then see here the x value hits the y on the graph

Answer 6

there is no real good way to graph it on here .... try wolframalpha.com if you need a good graphing utility

Answer 7

i have a graphing calc

Answer 8

it graphs it fine... but the numbers dont match up...

Answer 9

you want to graph the derivative of f(x)

Answer 10

once i do that what do i do?

Answer 11

you ... zoom in ... in the given values ... that are provided ... in the text

Answer 12

let me try that quick

Answer 13

its either that or there is a derivative function on your calculator somewheres

Answer 14

the first one is the easiest. the line tangent to
\[y=x^2\] at the point (0,0) is the y - axis and that has slope 0

Answer 15

once i graph the derivative, do I just see where all the f'(x) points hit the graph?

Answer 16

i would. yes

Answer 17

ok

Answer 18

let me go to wolfram alpha and graph it

Answer 19

here it is in the domain [0.49,0.51]to get an estimate of the slope at x = 1/2
http://www.wolframalpha.com/input/?i=y%3Dx^2++domain+.49..0.51

Answer 20

so i was suposed to get the slopes?

Answer 21

you can see that the slope of that "line" which is really a blow up of the curve, is about 1

Answer 22

ya

Answer 23

ok.. well now I get it, I was just trying to get teh actual y value and not the slope, lol

Answer 24

the actual y vaoue of the derivative IS the slope

Answer 25

you are being asked to estimate the slope of the line tangent to the curve at the various x values.

it is not a bad question, but the derivative of
\[y=x^2\] is
\[j'=2x\] so it is clearly easier to evaluate
\[f'(x)=2x\] at
\[x=0,x=.5,x=1, x=2\]
you will get
\[f'(0)=0,f'(.5)=1,f'(1)=2,f'(2)=4\]

Answer 26

but the question is asking you to estimate the slope of the tangent line. if you blow up the picture of
\[y=x^2\] around any one of those point, that is zoom in closely, you will get what looks like a line, and you can estimate the slope of the line from the picture.

Answer 27

ok thanks, now I understand it.