# `f(x) = sqrt(x^4 + cx^2)` Describe how the graph of `f`varies as `c` varies. Graph several members of the family to illustrate the trends that you discover. In particular, you should...

`f(x) = sqrt(x^4 + cx^2)` Describe how the graph of `f`varies as `c` varies. Graph several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and minimum points and inflection points move when `c` changes. You should also identify any transitional values of `c` at which the basic shape of the curve changes.

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### 1 Answer

We can factor the function to get `f(x)=\sqrt{x^2(x^2+c)}=|x|\sqrt{x^2+c}`. There are several cases.

When `c=0` , the domain is all real numbers, and the function becomes the simple quadratic `f(x)=x^2` . This has a minimum at `x=0` , and no inflection points, since the function is concave up for all real numbers.

The graph is given by

When `c>0` , the function is more complicated. The domain is still all real numbers, but now the derivative is found to be (by using the original, unfactored function):

`f'(x)=\frac{1}{2}(x^4+cx^2)^{-1/2}(4x^3+2cx)` factor numerator and denominator

`=\frac{x(2x^2+c)}{|x|\sqrt{x^2+c}}` replace `|x|` with its definition

`=\frac{2x^2+c}{\sqrt{x^2+c}}`

The graph has a minimum at `x=0` which we can see from the original function since it vanishes there and must be positive everywhere else.

The second derivative is `f''(x)=+-\frac{x(2x^2+3c)}{(x^2+c)^{3/2}}` . A sample graph is given by

When `c=-k<0` , we can carry out exactly the same computations as in the positive case, except that now there are x-intercepts at `x=+-\sqrt k` . But the domain is only when the argument is positive, which means that `x<=-k` and `x>=k` .

The first derivative is `f'(x)=\frac{2x^2-k}{\sqrt{x^2-k}}` , which indicates the minimums at the x-intercepts.

Finally, the second derivative is `f''(x)=+-\frac{x(2x^2-3k)}{(x^2-k)^{3/2}}` which has inflection points at `x=+-\sqrt{{3k}/2}` .