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Consider the functions $f(x)=\sqrt{5+4x-x^2}$ and $g(x)=|x-3|+2$. In what domain of $x$ the function $f(x)$ lie above the function $g(x)$, i.e. $f(x)>g(x)$?

I think I should subtract the function $f(x)$ from $g(x)$, find where the new function

$$\sqrt{5+4x-x^2}-(|x-3|+2)$$

is positive, but I'm not sure.

Infinity
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    The new function should be $\sqrt{5+4x-x^2} - |x-3| - 2$, other than that, you are on the right track. Keep going! – 5xum Sep 11 '14 at 12:52

2 Answers2

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Yeah your method is correct.

Since $f(x)\gt g(x)$

Subtract g(x) from both sides to get

$f(x)-g(x) \gt 0$

Then you have two choices. The first is you do the algebra (not my cup of tea).

The better choice is use your graphical calculator or wolfram alpha to plot the graph

$y=f(x)-g(x)$

Then you just look at the interval in which the graph is above zero (most graphical calculators can do this for you)

lagrange103
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First, obtain the interval where $f(x)$ is defined.

Then, split this interval, to dump the absolute values, i.e figure where $x-3$ is positive and when its negative.

Last, sum it all up

Snufsan
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