# `f(x) = (cx)/(1 + (c^2)(x^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) = (cx)/(1 + (c^2)(x^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.

### Textbook Question

Chapter 4, 4.6 - Problem 33 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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`f_c(x)` is defined everywhere and is infinitely differentiable.

`f_c(x) = g(cx)` where `g(x) = x/(1+x^2).`

So

`f'_c(x) = g'(cx)*c = c*(1-(cx)^2)/(1+(cx)^2)^2,`

`f''_c(x) = c^2*g''(cx) = c^2*2(cx)*((cx)^2-3)/(1+x^2)^3.`

The graphs for negative c's are the same as for positive c's except that x is replaced with -x:

`f_(-c)(x) = f_c(-x).`

Therefore we can consider only nonnegative c's.
Also, for c=0 `f_c` is identically zero.

For positive c:

`f'_c` is positive on `(-1/c,1/c)` and is negative outside. So f is increasing on `(-1/c,1/c)` and is decreasing on `(-oo,-1/c)` and on `(1/c,+oo).` And -1/c is a local minimum, 1/c is a local maximum.

`f''_c` is positive on `(sqrt(3)/c,+oo),` negative on `(0,sqrt(3)/c),` positive on `(-sqrt(3)/c,0)` and negative on `(-oo,-sqrt(3)/c).` Correspondently f is concave upward on `(sqrt(3)/c,+oo)` and on `(-sqrt(3)/c,0),` and is concave downward on `(-oo,-sqrt(3)/c)` and on `(0,sqrt(3)/c).`

So 0 and `+-sqrt(x)/c` are the inflection points.

Please look at the graphs here: https://www.desmos.com/calculator/lbdznlp1c2

Red graphs are for negative c's, green for positive and blue for c=0.