I understand the restricted domains of inverse trig functions, but what about:
I don't quite understand how to find the domain and range of this function.
3 Answers
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For the domain, we know that in $\cos^{-1}(t)$, we must have $-1 \leq t \leq 1$. So in our given expression we need $-1 \leq 3x-4 \leq 1$. We can solve this for $x$: $$ -1 \leq 3x-4 \leq 1 \\ 3 \leq 3x \leq 5 \\ 1 \leq x \leq \frac{5}{3} \\ $$
Similarly for the range, $\cos^{-1}(t)$ returns a number between $0$ and $\pi$ for whatever $t$ is. We can use this to find the range of the above expression (just put $3x-4$ in place of $t$): $$ 0 \leq \cos^{-1}(3x-4) \leq \pi \\ 0 \leq 3\cos^{-1}(3x-4) \leq 3\pi \\ 2 \leq 3\cos^{-1}(3x-4)+2 \leq 3\pi+2 $$
answered Nov 12, 2014 at 2:23
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$\begingroup$ That makes a lot of sense thank you. I'm confused on the range. If normally it is 0 < 3y + 2 < pi, I don't see how you got 2 < y < 3pi + 2 $\endgroup$– boopNov 12, 2014 at 2:36
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$\begingroup$ Let's denote $\cos^{-1}$ as $f$. The range of $f$ is $[0,\pi]$. But we don't need the range of $f$: we need the range of $3f+2$, so in order to get the new range, we need to do the same operations on the endpoints of the range that are acting on the function. Namely we need to take the endpoints of the range, multiply them by $3$ and then add $2$. $\endgroup$ Nov 12, 2014 at 2:41
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Domain = $\{x \colon -1 \leq 3x - 4 \leq 1 \}$
Range = $\{3y+2 \colon 0\leq y\leq \pi\}$
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To find the domain of the function, you should then be able to solve $3x-4=1 $ and $3x-4 =1$. Therefore for the minimum x value, $x=1$ and the maximum value is equal to $x=5/3$. The range is the maximum y value given the domain in this case.
