11:43
In this case, the Lagrance bound is actually off by something like a factor of 2, and certainly isn't good enough to get the answer required to 3 decimal places. So I think they just want you to evaluate the error directly.
So for any given x you can find that just by computing x e^(0.2x) and x + x^2/5 + etc. and taking the difference.
The worst error is almost certainly going to be at the end of the range where the function is largest, increasing fastest, etc. (The worst error in a truncated Taylor series is almost always at the end of the range.)
For a multiple-choice question, you can just calculate the error at each of the x values they give and see if one of them produces an error very close to 0.01.
Oh, in this particular case you don't need to rely on any heuristic "worst error is usually at largest |x-a|" stuff, because the error is just the rest of the Taylor series for x e^(x/5), all of whose coeffs will be positive, so obviously larger x will make it larger. So the biggest error is absolutely definitely at the biggest x.
So, anyway, for a not-multiple-choice question: guess a value of x, see what the error is; if the error's too large, try a substantially smaller value of x, if it's too small, try a substantially larger value of x, repeat until you've got one x where the error is too big and one where it's too small.
Then do a binary search: pick a value somewhere near the middle of the range (literal binary search always uses exactly the middle, but you don't have to do that; pick a value with fewer digits so the calculations are easier, pick a value nearer one end if it seems like the error is nearer the target at that end than at the other end, etc.) and evaluate the error, and now pairing that with one of the two values you had before gives you a new smaller interval where the error is [... continues]
... too small at one end and too big at the other.
So e.g. maybe you start with x=1 and find the error is about 2.8x10^-6. Much too small, try a larger x.
x=2: error is about 0.00018, still too small but getting nearer the mark. Maybe don't double it this time; try x=3.
x=3: error is about 0.002. Damn, should have gone higher after all. x=4 next?
x=4: error is about 0.013. Aha, so somewhere between 3 and 4, probably quite close to 4 because 0.013 isn't much bigger than 0.01. So try somewhere in between, maybe nearer to 4. Let's try 3.8 and see what we get.
x=3.8: error is about 0.0092. Nice. So we're looking for a value between 3.8 and 4, probably nearer 3.8 because 0.0092 is nearer 0.01 than 0.013 is. Literal bisection would try 3.9 next but let's try 3.85 instead.
x=3.85: error is about 0.00993. Getting close. Now we're looking for a value between 3.85 and 4, probably much nearer 3.85. Let's try bumping it up to 3.86 and see what we get.
x=3.86: error is about 0.01009. Also close but on the other side, so now we need a value of x between 3.85 and 3.86. Let's try 3.855.
x=3.85: error is about 0.010014. Still too large; need something between 3.85 and 3.855. The error at 3.855 is quite a bit nearer the mark than the error at 3.850, so let's try something closeish to that. Maybe 3.853?
x=3.853: error is about 0.00998. So we need x between 3.853 and 3.854. Might as well try the midpoint at 3.854.
x=3.854: error is about 0.009998. Too small but very close. So for an error of 0.01 we need something between 3.854 and 3.855, probably somewhat nearer 3.854. If we want to be sure of getting the right value rounded to 3dp, we need to know which side of 3.8545 the correct value is on. I bet it's the low side, but let's check.
x=3.8545: error is about 0.010006. Too large. So the place where we cross 0.01 is between 3.8540 and 3.8545, so to 3dp it must be 3.854. Done.
That's pretty laborious when you write everything out, but you can do it in a minute or two if you aren't so verbose :-).
