1 Answer | Add Yours
To start, we must take note that this function may not have defined extreme values on the given interval because the interval is open. In order for us to have a maximum, `f_max`, the following conditions must hold true:
These conditions exist because the fact that the interval is open indicates that `f(0)` and `f(4)` are not in the interval and cannot be considered for extreme values. Similarly for the minimum, `f_min`, to exist, the following conditions must be met:
1) `f_min < f(0)`
So, let's start by evaluating `f(0)` and `f(4)` to find the numbers with which `f_min` and `f_max` must be compared.
`f(0) = e^0/(1+e^0) = 1/(1+1) = 1/2`
`f(4) = e^4/(1+e^4) ~~ 0.982`
Now, we must find possible candidates for extreme values in the given interval. To do this, let's take the derivative of the function and set it to zero to see whether any local maxima or minima are between 0 and 4. Here, we will use the quotient rule.
`(df)/(dx) = (e^x(1+e^x) - e^x*e^x)/(1+e^x)^2`
Simplifying through the distributive property and subtraction:
`(df)/(dx) = (e^x + e^2x - e^2x)/(1+e^x)^2 = e^x/(1+e^x)^2`
We have a problem in terms of finding locations where this derivative is equal to zero: it never can! Remember, when trying to find where a function like this can be zero, all we need to do is look at the numerator. An exponential function like this can never be zero. It can only approach zero as `x` approaches `-oo`.
As a result, we have no candidates on this interval to be an extreme value.
Therefore, considering there is no point at which the derivative of the function is zero and considering that the interval is open, there are no extreme values in this case.
See the link below for some more information on these sorts of problems.
I hope that helps!
We’ve answered 319,832 questions. We can answer yours, too.Ask a question