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) · (x2 + 4)1/2 + (x) · (2x) · (1/2) · (x2 + 4)-1/2
= (x2 + 4)1/2 + x · (x2 + 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) · (x2 + 4)-1/2 + (2x) · (x2 + 4)-1/2 - (x2) · (2x) · (1/2) · (x2 + 4)-3/4
= (3x) · (x2 + 4)-1/2 - (x3) · (x2 + 4)-3/4
We can simplify this one step further by finding a common denominator. This will result in the following
f ''(x) = (2x3 + 12x) · (x2 + 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)(x2 + 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 x2 + 6 = 0, we get imaginary solutions (try it with quadratic formula). Now we can consider the inflection point theorem that says:
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
f ''(1) = 14 · (5-3/2)> 0
Since 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).