James B. answered • 05/29/15

PhD student at the University of Minnesota with teaching exp

Hi Adam,

Let us begin your problem by finding the second derivative of the given function. The first step is to find

*f'(x)*using the product rule. This is given by*f'(x) = (1) · (x*

^{2}+ 4)^{1/2 }+ (x) · (2x) · (1/2) · (x^{2}+ 4)^{-1/2}

*= (x*

^{2}+ 4)^{1/2}+ x · (x^{2}+ 4)^{-1/2}At this point, notice that this function is continuous and always positive (this fact will be useful later). Next, we take another derivative giving us

*f ''(x).**f ''(x) = (1/2) · (2x) · (x*

^{2}+ 4)^{-}^{1/2}+ (2x) · (x^{2 }+ 4)^{-1/2 }- (x^{2}) · (2x) · (1/2) · (x^{2}+ 4)^{-3/4}

*= (3x) · (x*

^{2}+ 4)^{-1/2}- (x^{3}) · (x^{2}+ 4)^{-3/4}We can simplify this one step further by finding a common denominator. This will result in the following

*f ''(x) = (2x*

^{3}+ 12x) · (x^{2}+ 4)^{-3/4}Now is a good point to notice that the denominator of this expression is always positive (so we don't have any divide by zero situations) and the numerator is just a degree 3 polynomial, therefore the function is continuous. Next we notice that the numerator is a degree three polynomial that can be factored as

*(2x)(x*

^{2}+ 6)If we set this equal to zero (to find the points of inflection), then we get only one solution. This solution comes from solving

*2x = 0*. This solution is simply*x = 0*

Why no other solutions? Well if we try to solve

*x*we get imaginary solutions (try it with quadratic formula). Now we can consider the inflection point theorem that says:^{2 }+ 6 = 0,*If f'(x) exists and f ''(x) changes sign at a point x=c, then the point (c , f(c)) is an inflection point. Further, if f ''(x) also exists, it must be equal to zero.*

Lets take inventory of what we have done learned about this function so far

*f'(x)*is continuous so it exists everywhere.*f ''(x)*is continuous so it exists everywhere.*f ''(0) = 0*

Now if we choose test point

*x = -1*and*x = 1*. We have*f ''(-1) = -14 · (5*

^{-3/2})< 0*and*

*f ''(1) = 14 · (5*

^{-}^{3/2})> 0Since

*f ''(x)*changes sign at*x=0*, we can conclude that this corresponds to an inflection point. Therefore our function is- concave down from -∞ to 0,
- concave up from 0 to ∞,
- it has an inflection point at (0,0),
- and it does not have a maximum or minimum because it is always decreasing (since the derivative is always positive).