# f(x) = x/(x^3 + x^2 + 1) Produce graphs of f that reveal all the important aspects of the curve. In particular, you should use graphs of f' and f''

### Textbook Question

Chapter 4, 4.6 - Problem 5 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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lfryerda | High School Teacher | (Level 2) Educator

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The denominator of the function vanishes at x\approx -1.465571 which means there is a vertical asymptote there.  Since the numerator is of degree one and the denominator is degree three, there is a horizontal asymptote at y=0.

The first derivative is:

f'(x) = \frac{x^3+x^2+1-x(3x^2+2x)}{(x^3+x^2+1)^2}   simplify the numerator

=-\frac{2x^3+x^2-1}{(x^3+x^2-1)^2}

which has a zero at x\approx 0.6573 .  This indicates a local extrema at that location.

The second derivative is

f''(x)=\frac{2x(3x^4+3x^3+x^2-6x-3)}{(x^3+x^2+1)^3}

which has a zero at x=0, x\approx -0.48998 and x\approx 1.100204

The first derivative graph looks like

and the second derivative graph looks like:

We can see from both the first and second derivative that the function has a vertical asymptote where indicated before.  In addition, the roots of the second derivative indicate corresponding inflection points in the original function.  Finally, the zero of the first derivative, combined with the second derivative being negative, indicates that the original has a local maximum there.  These features with the zero of the function at the origin give the final graph of: