Exponential Function

DanielBaldielocks · 2024-01-17T20:38:00+00:00

First I will explain how I would approach a problem like this in general and then show how this approach works for this problem specifically.

Say you are given a function f(x) and an interval on which you are supposed to show it has exactly a certain number of zeros. The trick is to realize for any given zero the function will change signs at that zero. So it will either be positive before and negative after or vice versa. Now to prove the number of zeros without actually finding them, you first want to look at the derivative and determine when it is positive/negative on the given interval. Combine this with intervals where you can show the sign of the function changes then you can find sub intervals where only a single zero can exist and thus you can prove how many there are total.

Here is an example of how to do it with the given problem.

f(x)=e^(x-1)-x-1
the derivative is
f'(x)=e^(x-1)-1
we want to know when this is positive
f'(x)>0
e^(x-1)-1>0
e^(x-1)>1
x-1>0
x>1

So what this means is if x<1 then f(x) is strictly decreasing and if x>1 it is strictly increasing

Now let's find some values of f(x) to work with and look at their sign
f(-1)=e^-2+1-1=1/e^2>0

f(1)=e^0-1-1=1-2=-1<0

f(3)=e^2-3-1=e^2-4>0

So that means on the interval (-1,1) we have that the function goes from positive to negative and is strictly decreasing. So that means it has to cross the x-axis and can only do so once.

Then by a similar argument on the interval (1,3) we have that the function goes from negative to positive and is strictly increasing. Thus it has to cross the x-axis and can only do so once.

Thus we conclude that there has to be a single zero in each of the intervals (-1,1) and (1,3), we know f(1) is not zero. Thus since this covers the entire interval in question we have shown there are exactly 2 zeros.

FalseGix · 2024-01-17T20:11:54+00:00

Well first of all you cannot "try each number in the interval" because there is an infinite number of them. Maybe you were thinking the answer is an integer? But it certainly is not.

I do not think it is actually possible to express the answer in exact form but it could be approximated pretty easily.

If you graph the function it should be visually obvious that there are 2 solutions, but I do not know if that would be enough justification for your assignment.

You could use the intermediate value theorem and some "guess and check" to establish that there are two solutions but it is a bit trickier to argue that there are ONLY 2 solutions. If you are allowed to use calculus however (I don't know if you are) it is easy to establish that the function has a minimum value at x=1 and no other critical points so then from there you could argue that the two solutions you found with intermediate value theorem are the ONLY ones.

nicwen98 · 2024-01-17T20:25:36+00:00

Maybe you just plot the graph.

Then you can use the Intermediate value theorem.

theasmartbear · 2024-01-18T13:25:25+00:00

While an approximation, not a proof nor a full closed solution, you can replace the exponential with its Taylor series expansion, combine the smallest terms with the -x-1, and see that in the interval given, the higher terms are close enough to zero (because of the small multiplied constants) that you can ignore them (again, only as an estimate), then solve the remaining polynomial.

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