SolutionFor both of these, itโs sufficient to consider only sin(x)^sin(x). It stands to reason that the maximums will happen when sin(x)^sin(x) is at its maximum value, and its minimums will happen when sin(x)^sin(x) is at its minimum value. The x^ part really just makes the function prettier to look at. The function lacks any continuity when sin(x) is negative, so we can consider just the regions where sin(x) is positive.
For the maximums, this is very easy: The largest sin(x) can be is 1, and so the largest value for sin(x)^sin(x) is also 1. So the monotonically increasing function which passes through each local maximum is y = x^1, or just y = x.
For the minimums itโs a bit trickier: The smallest sin(x) can be is 0, and while 0^0 is indeterminate, it happens to approach 1 in this case - our minimum is somewhere else (side note: sin(x)^sin(x) approaching 1 for these values is why our max function also exactly bounds the flailing arms of the function). So we turn to derivatives. We can do this using either sin(x)^sin(x) or x^x - I chose to use sin(x)^sin(x) here and x^x (sort of) in the extra credit. Using sin(x)^sin(x) here has the added benefit of also showing the maximums are when sin(x) = 1.
y = sin(x)^sin(x)
y = e^(ln(sin(x)) * sin(x))
yโ = sin(x)^sin(x) * (cos(x)/sin(x) * sin(x) + ln(sin(x)) * cos(x)) โ Chain rule, product rule, and chain rule again
yโ = sin(x)^sin(x) * (cos(x) + ln(sin(x)) * cos(x)) โ Simplify
yโ = sin(x)^sin(x) * cos(x) * (1 + ln(sin(x))) โ Factor
We can set each of these factors to 0.
sin(x)^sin(x) will never be 0.
cos(x) will be 0 when x = ฯ/2 + 2ฯn (2ฯn instead of ฯn because weโre skipping over the solutions where sin(x) is negative). These are the maximums, because when x = ฯ/2 + 2ฯn we have sin(x) = 1.
1 + ln(sin(x)) = 0
ln(sin(x)) = -1
sin(x) = e^-1 = 1/e
And thatโs actually sufficient for our purposes - we could of course say x = arcsin(1/e), but our goal is to calcualte the value of sin(x)^sin(x). Since sin(x) = 1/e, we have sin(x)^sin(x) = (1/e)^(1/e). So our monotonically increasing function which hits all the minimums is y = x^((1/e)^(1/e)), or: y = (e-th root of e)-th root of x.
Graphed solution visible
here.
For the extra credit, we can do the same basic idea: Find the maximum of |sin(x)|^sin(x). Since we know weโre in a region where sin(x) is negative, we can model this as x^-x, and the solution will be valid as long as x is something between 0 and 1, within sin(x)โs positive range.
y = x^-x = e^(ln(x) * -x)
yโ = x^-x * (-1 - ln(x))
x^-x wonโt ever be 0, and -1 - ln(x) will be 0 when ln(x) = -1, or x = 1/e again, same as before. So our maximum value happens at (1/e)^(-1/e), which simplifies to e^(1/e). In other words, our new maximum function is x^(e^(1/e)), or x to the power of the e-th root of e.
Graph of all three solution functions
And, finally - I mentioned that we should also be able to expand the function to its less-defined regions using y = x^-|sin(x)|^sin(x). I donโt care enough to post the full work here, but the reciprocal of the three answer solutions above are what works here: y = 1/x, y = 1/x^((1/e)^(1/e)), and y = 1/x^(e^(1/e)).
Graph. Note that the upper function here only applies in regions where the original function is already defined, and so the other parts of this extension are invalid - you can toggle the main function at the top off to see only the parts that truly apply.
Nothing particularly interesting happens if you extend things to values of x less than 0 - the graphs (more or less) just reflect across the y-axis.
Bonus graph with everything + phase shifts and pretty colors
