`f(x) = ln(x^2 + c)` 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...

`f(x) = ln(x^2 + c)` 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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Textbook Question

Chapter 4, 4.6 - Problem 32 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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For c>0 f(x) is defined and infinitely differentiable everywhere. For c=0 f is undefined at x=0, and for c<0 f is undefined for x<=sqrt(-c) and x>+-sqrt(-c).

Find f' an f'':

`f'_c(x) = (2x)/(x^2+c),`

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

1. For c<0:
f' is negative for `xlt-sqrt(-c),` f decreases. f' is positive for `xgtsqrt(-c),` f increases.
f'' is always negative, f is concave downward on `(-oo,-sqrt(-c))` and on `(sqrt(-c),+oo).`

2. For c=0: (`f(x)=2ln|x|` )
`f'(x)=2/x` is negative for x<0, f decreases, and positive for x>0, f increases.
`f''(x) = -2/x^2 ` is always negative, f is concave downward on `(-oo,0)` and on `(0,+oo).`

3. For c>0:
f' is negative for c<0, f decreases, f' is positive for x>0, f increases. At x=0 f'(x)=0 and this is a minimum.
f'' has roots at `x=+-sqrt(c).`
F'' is negative for `xlt-sqrt(c)` and f is concave downward,
F'' is positive for x on `(-sqrt(c),+sqrt(c))` and f is concave upward,
F'' is negative for `xgtsqrt(c)` and f is concave downward.
So `x=+-sqrt(c)` are inflection points.

The only transitional value of c is zero.
Please look at the graph here: https://www.desmos.com/calculator/89vayoctfv
(the green graphs are for c>0, red for c<0 and blue for c=0)

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