`f(x) = xsqrt(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) = xsqrt(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.

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Textbook Question

Chapter 4, 4.6 - Problem 30 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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This function is defined only on [-c,c] (we can assume that c is positive) and infinitely differentiable inside the interval.

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

`f''_c(x) = (-4x*sqrt(c^2-x^2)-(c^2-2x^2)*(-2x)/(2sqrt(c^2-x^2)))/(c^2-x^2) = (x*(2x^2-3c^2))/(c^2-x^2)^(3/2).`

f' is negative for x in `(-c,-c/sqrt(2))` and in `(c/sqrt(2),c),` f decreases there, and f' is positive for x in `(-c/sqrt(2),c/sqrt(2)),` f increases.

So `-c/sqrt(2)` is a local minimum, `c/sqrt(2)` is a local maximum.

About f'': `2x^2-3c^2` is negative on (-c,c) so f'' is positive on (-c,0) and is negative on (0,c). f is concave upward on (-c,0) and is concave downward on (0,c). So x=0 is an inflection point.

The general shape of the graph is the same for all c's.

Please look at the graphs here: https://www.desmos.com/calculator/4a0q7l7sec

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