Question 1043530
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If {{{p(x)}}} = {{{2sqrt(x-5)+3x}}}, what is the minimum value of P?
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The square root part domain is x >= 5.
So, it is the domain for the entire p(x).


In this domain {{{sqrt(x-5)}}} monotonically grows.
3x monotonically grows too.


Hence, p(x) monotonically grows.
It implies that the minimum value of p(x) is at x = 5. You can easily calculate it.

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{{{graph( 330, 330, -2.5, 10.5, -2.5, 25.5,
          2*sqrt(x-5)+3x
)}}}


<B>Figure</B>. Plot f(x) = {{{2sqrt(x-5)+3x}}}

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