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How to find absolute extrema of this function?
Not asking for anyone to do whole problem as I have no idea how long this will be to find, just no idea where to start or what to do..
Let g(t) = (2t-1)e^(-1) be a function
1) Find the absolute extrema of g(t) on the interval [0,Infinity)
Any advice or where to go from there would be really appreciated, thank you.
2 Answers
- JamesLv 77 years ago
An absolute extremum of g(t) exists at a point t if and only if all values of g(t) are greater than (in the case of an absolute minimum) or less than (for a maximum).
A good place to start is to get a mental image (or heck, a computer-generated graph) of the function. You know that (2t-1) is a line with no absolute extrema, and you know that e^(-1) is a constant, so your function is a line with a slope of 2/e. Pick any point on that line, and you can demonstrate that a point to the left or right will have a greater or lesser y-coordinate, and therefore there are no absolute extrema.
- 7 years ago
f(t) = (2t - 1)e^(-1)
f ' (t) = 2te^(-1) - e^(-1)
f ' (t) = 2e^(-1) > 0
Thus, f is increasing. Thus, on [0, inf) we see that f(0) => f(t) for all t in [0, inf). Thus, f(0) = (2*0-1)e^(-1) = -1/e is the minimum of f and there is no maximum.
