Help Needed: Differentiation in Maths - Find F(x) & t(x)

[VooDoo]

We did this new mathematics thing in class today which I did not understand, it came under the topic of differentiation.

Firstly we were given the equation

f(x)=E^x

Then find the value of F(x) for x=0
F(0)=E^0
F(0)=1
therefore (0,1)

Then it said find the equation of the tangent at x=0

Which I worked out to be Y(tangent)=X+1 [I will call this equation t(X)]

Then it said draw up a table ranging from -.2 to .2 with increments of 0.05 for the equation

[f(x)-t(x)/f(x)]*100

Now my question is what the hell are we finding here??

Thanks in advance for any help.

 

[dextercioby]

Who's t(x)...?

Daniel.

 

[HallsofIvy]

First of all, it is "e", not "E". Mathematically, small and capital letters may represent very different things.

Second: "Then it said draw up a table ranging from -.2 to .2 with increments of 0.05 for the equation

[f(x)-t(x)/f(x)]*100

Now my question is what the hell are we finding here??"

My question is "what the hell are you talking about??"
You haven't told us what t(x) is! I suspect that you meant t(x) to be x+ 1, the tangent line. In that case, you are finding the percentage error in using the tangent line to approximate e^x around x=0.

 

[VooDoo]

First of all, it is "e", not "E". Mathematically, small and capital letters may represent very different things.

Second: "Then it said draw up a table ranging from -.2 to .2 with increments of 0.05 for the equation

[f(x)-t(x)/f(x)]*100

Now my question is what the hell are we finding here??"

My question is "what the hell are you talking about??"
You haven't told us what t(x) is! I suspect that you meant t(x) to be x+ 1, the tangent line. In that case, you are finding the percentage error in using the tangent line to approximate e^x around x=0.

Thanks for that, just what I was after. I mentioned what t(x) was but I agree it was not clear enough.

Hmm, HallsofIvy do you know any alternatives to this method?

 

[Corneo]

Differentials can approximate functions at a specific point.

$$\Delta y \approx f'(x) \Delta x$$

The change in a function at near point $f(x)$ is approximately the numerical derivative at x multiplied by the small change of x.

 

[arildno]

Here is what I guess you were supposed to be doing:
1) Let $$f(x)=e^{x}$$
2) The best linear approximation to f(x) at x=0 is given by $$t(x)=f(0)+f'(0)(x-0)=x+1$$
this is also called the tangent line to f at x=0
3) You are now to find the PERCENTWISE RELATIVE ERROR E(x) between f(x) and t(x) at the interval given:
$$E(x)=(\frac{f(x)-t(x)}{f(x)})*100$$