How to sketch $y = \sqrt{x-1} + \sqrt{6-x}$

Question

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.

Answer 1

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!