Question

Use differentials and the graph of f to approximate (a) f(1.9) and (b) f(2.04).

How do I approach this problem and go about doing it?

Answer 1

supposing that you have a given function y=f(x).
The derivative of this is

\[\frac{ dy }{ dx }=f '(x)\] Please multiply both sides of this new equation by dx and simplify the result. Type your answer here.

Answer 2

the resulting equation is that for "differential approximation" of function y=f(x).

Answer 3

Thanks. Where does the graph help me though?

Answer 4

I've been reviewing the concept of "differential" with you. Mind writing out the formula for the differential of y, based upon \[\frac{ dy }{ dx }=f '(x)\]...???

Answer 5

dy= f'(x)dx?

Answer 6

good. Here, dy represents the error in the output if the error in the input is dx. Seen this before?

Answer 7

Yes, I think I have

Answer 8

I would prefer to use this equation for "linear approximation:"\[y=y _{0}+f '(x _{0})(x-x _{0}).\] If you look carefull at the rightmost term, you'll see that it 's just about the same thing that we got for a differential, above. Which approach would you like to use to solve the problem at hand?

Answer 9

I am familiar with the linear approximation way

Answer 10

Great. Now let's address finding an approximation for f(1.9). to find this approx. value of f)x) at x=1.9, we need the following data:

1) the slope of the tangent line to the curve at x=2. Can you find that from the graph?
2) The value of the function (the graphed function) at x=2.
3) the difference between 1.9 and 2.0.

Find or calculate these, please.

Answer 11

1) I think the slope of the tangent line is 1
2) At x=2, the function value is 1
3) the difference is 0.1

Answer 12

As you move from left to right, does your tangent line go up or down?
If you begin at x=2, are you increasing or decreasing x if you move to x=1.9?

Answer 13

The tangent line goes down and from x=2, I'd be decreasing x if I move to x=1.9

Answer 14

if the tangent line "goes down," wouldn't that imply that the slope of your tangent line is (-), instead of (+)?

Answer 15

If x is decreasing, as it is here, and you're moving from x=2 to x=1.9, what is dx (the change in x)?

Answer 16

Oh, yes that's true. So my slope would be negative instead of positive. And would the change in x be -0.1 instead of 0.1?

Answer 17

Right on.

Time to put all of this together. The "linear approx." formula is \[y=y _{0}+f '(x _{0})(x-x _{0}).\]Determine whether you can now recognize all the info you have to substitute into this formula to obtain an approximation for f(1.9).

Answer 18

\[f(1.9)~is~approximately~what?\]

Answer 19

f(1.9)= 1+ (-1)(2-0.1)?

Answer 20

-.9?

Answer 21

Comment #1: You've said, correctly, that the slope of the tangent line at x=2 is -1. But your latest result doesn't reflect that.
Comment: #2: cx = 1.9-2.0=-0.1 (not -0.9).

Try again? f(1.9) is approx. equal to .... ???

Answer 22

f(1.9)= 1+ (1)(-0.1)=0.9?

Answer 23

Did I plug in the numbers incorrectly?

Answer 24

@mathmale

Answer 25

Yes. \[y=y _{0}+f '(x _{0})(x-x _{0})\]is your starting point.

Note that y-sub-0 is 1.
Note that the slope, m, f'(x-sub-0) is -1 (not +1).
Note: x-x-sub-0 is -0.1.

try again, please. y = ??

Answer 26

f(1.9)= (1)+(-1)(-0.1)=1.1

Answer 27

that looks good! Just supposing we begin at (2,1) and move along the curve to the left. Will y increase or decrease? Does your answer fit in with your estimated function value, 1.1? What's your final conclusion?

Answer 28

Then y would increase if we moved along the curve to the left instead of the right. Which means that it fits the estimate because if x was to increase to 2, then the y would decrease again to 1. Right?

Answer 29

Estimating y for x=2.04 is not much different. As before, x0 (meaning 'x sub 0' is 2; y0 is 1. The slope is -1 as before. Only (x-x-sub-0) is different. Care to give that problem a try? Would you expect the new y value to be larger or smaller than 1? Why?

Unfortunately, I need to get off the Internet now. But I'll be back on later.


To answer your most recent question, YES!

Answer 30

Okay, thank you so much for your help! I think I can handle part b on my own now. Thanks again!!

Answer 31

My great pleasure. Good luck!