What are the critical values of #f(x)=1/sqrt(x^2+4)-ln(x^2+4)#?

To find the critical values of ( f(x) = \frac{1}{\sqrt{x^2 + 4}} - \ln(x^2 + 4) ), we need to first find its derivative, set it equal to zero, and then solve for ( x ).

[ f(x) = \frac{1}{\sqrt{x^2 + 4}} - \ln(x^2 + 4) ]

[ f'(x) = -\frac{1}{2(x^2 + 4)^{\frac{3}{2}}} \cdot 2x + \frac{2x}{x^2 + 4} ]

[ f'(x) = -\frac{x}{(x^2 + 4)^{\frac{3}{2}}} + \frac{2x}{x^2 + 4} ]

Now, set ( f'(x) ) equal to zero:

[ -\frac{x}{(x^2 + 4)^{\frac{3}{2}}} + \frac{2x}{x^2 + 4} = 0 ]

[ -x(x^2 + 4)^{-\frac{3}{2}} + 2x(x^2 + 4)^{-1} = 0 ]

[ -x + 2(x^2 + 4) = 0 ]

[ -x + 2x^2 + 8 = 0 ]

[ 2x^2 - x + 8 = 0 ]

This is a quadratic equation. Solve it for ( x ) using the quadratic formula:

[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

Where ( a = 2 ), ( b = -1 ), and ( c = 8 ).

[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot 8}}{2 \cdot 2} ]

[ x = \frac{1 \pm \sqrt{1 - 64}}{4} ]

[ x = \frac{1 \pm \sqrt{-63}}{4} ]

Since the square root of a negative number is not real, there are no real critical values for this function.