Question 1195991
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Find the range of {{{f(x) = (x-3)/(sqrt(3-sqrt(x+2)))}}}<br>
(1) The numerator is monotonically increasing;
(2) The denominator is monotonically decreasing<br>
Therefore, the function is monotonically increasing.<br>
Because of the {{{sqrt(x+2)}}} in the denominator, the minimum value for x is -2.  Then, since the function is monotonically increasing, the minimum value of the function is {{{f(-2)=(-5)/sqrt(3)}}} = -2.88675 to 5 decimal places.<br>
The denominator of the function is positive, because it is of the form sqrt(A).  The value of x that makes the denominator 0 is x=7; so x must be less than 7.<br>
But as x approaches 7 from the left, the denominator gets arbitrarily close to zero; and that means the value of the function gets arbitrarily large.<br>
So there is no maximum value of the function.<br>
ANSWER: The range of the function is from {{{-5/sqrt(3)}}} to infinity.<br>
Here is a graph of the function on the window [-3,8,-5,50], showing the minimum value of the function at (-2,-2.88675).  Note that the graph is so steep close to x=7 that the graphing utility can't graph the function past a certain point, making it appear that the function has a maximum value.<br>
{{{graph(600,400,-3,8,-5,50,(x-3)/(sqrt(3-sqrt(x+2))))}}}<br>
But here is a graph of the function on the window [6.99,7,0,1000], showing that the function value is still increasing.<br>
{{{graph(600,400,6.99,7,0,1000,(x-3)/(sqrt(3-sqrt(x+2))))}}}<br>