How to sketch $y = \sqrt{x-1} + \sqrt{6-x}$
My solution:
It does not have any roots
Domain = [1,6]
Increasing till 3.5 and then decreasing
How to go on further?
Please help.
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1 Answers
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Nice work: what you've found is surely key information.
I'd add what $f(3.5)$ is equal to, as that's the maximum: $f(3.5) = 2\sqrt{\frac 52}= \sqrt{10}$.
And I'd find the values of $f(x)$ at the endpoints of the domain, i.e., the minima at $f(1) = f(6) =\sqrt{5}$.
And that gives us a function $y$ whose range is $\sqrt 5\leq y \leq \sqrt{10}$
That should give you plenty to work with when sketching the graph!
amWhy
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how will we find whether curve is concave or convex.Please show your calculations also since I am a novice. – user2369284 Dec 24 '13 at 18:06
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For more details of the shape near the endpoints, consider that near $1$ (say for $x\in[1,1.1]$), $\sqrt{x-1}$ grows very steeply and $\sqrt{6-x}$ is close to constant. So near $x=1$, the function is close to $\sqrt{x-1}+\mbox{constant}$. If you know what that would look like, it helps. – 2'5 9'2 Dec 24 '13 at 18:08
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The calculations are merely evaluating $y$ at the given values of $x$, nothing more than substitution. Note also that the function is symmetric about the line $x = 3.5$ – amWhy Dec 24 '13 at 18:09
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@amWhy I wanted to ask for the calculation of convex or concave not the one in your answer. – user2369284 Dec 24 '13 at 18:14
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It is not convex: it is concave "down" and you can use any number of tests to show this, including the second derivative test. What test do you know? – amWhy Dec 24 '13 at 18:20
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@amWhy don't know very clearly about 2nd derivative test. – user2369284 Dec 24 '13 at 18:22
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@amWhy: Needs another TU +1 – Amzoti Dec 25 '13 at 00:37
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@amWhy: Hello my friend. Maybe it is soon to say, but Happy Xmas. :) – Mikasa Dec 25 '13 at 02:30
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Nice feedback!+1 – Dec 25 '13 at 16:49