#can someone explain how to answer this question fully
22 messages · Page 1 of 1 (latest)
For the first one you can rewrite the sum of trig expressions as a single expression as shown in the image.
for the rest of the questions, think about what properties a function needs for it to have an inverse.
Yea what I extremely struggle with is like, how to do the range and domain of those types of questions
Like part iv as well I don’t know how to do that and I don’t know how to do the range of v)
As you may know domain is the values you evaluate the function on, in this case for $f$ the domain is $0 \leq x \leq 2\pi$. The range is the values you can get from evaluating the function over the domain.
Crystopher
It may sound hard at first since you may ask: How do I evaluate over every value in the domain if there are infinitely many of these?
Here's when you use some properties of functions like $f$ and $g$, more specifically they are continuous.
Crystopher
since they are continuous we can use the intermediate value theorem.
So now instead of finding every single evaluation of $f$ to find its range we can just find its lower and upper bound. Then by intermediate value theorem all other values in-between those must also be part of the range.
Crystopher
In short, find the lowest (call it $\alpha$) and highest values (call it $\beta$) that $f(x)$ can have in the interval $x\in[0,2\pi]$, then its range is $y \in[\alpha,\beta]$. For this you can use derivatives or properties of trigonometric functions.
Crystopher
Not necessarily, the ends of the domain are good to study but the highest and lowest value of the function may lie somewhere in-between.
For instance, you can see that
$f(0)=f(2\pi)=2\sqrt{3}$ but $f(\frac{\pi}{2})= 2$ and $\frac{\pi}{2}>0$
Crystopher
Meaning that here $f(0)$ is not the lower bound of the range since there is a smaller value, and there are probably even smaller values than that.
Crystopher