Single Variable Calculus, Chapter 4, 4.6, Section 4.6, Problem 4

Use the graphs of $f'$ and $f''$ to estimate the intervals of increase and decrease, extreme values, intervals of concavity and inflection points. Suppose that $\displaystyle f(x) = \frac{x^2 - 1}{40x^3 + x + 1}$.

If $f(x)= \displaystyle \frac{x^2 - 1}{40x^3 + x + 1} $, then by using Quotient Rule..


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{(40x^3 + x + 1) (2x) - (x^2 - 1)(120x^2 + 1) }{(40x^3 + x + 1)^2}
\\
\\
f'(x) =& \frac{80x^4 + 2x^2 + 2x - 120x^4 - x^2 + 120x^2 + 1}{(40x^3 + x + 1)^2}
\\
\\
f'(x) =& \frac{-40x^4 + 121x^2 + 2x + 1}{(40x^3 + x + 1)^2}


\end{aligned}
\end{equation}
$


Again, by using Quotient Rule and Chain Rule,

$\displaystyle f''(x) = \frac{(40x^3 + x + 1)^2 (-160x^3 + 242x + 2) - (-40x^3 + 121x^2 + 2x + 1) (2(40x^3 + x + 1)) (120x^2 + 1) }{[(40x^3 + x + 1)^2]^2}$

which can be simplified as

$\displaystyle f''(x) = \frac{80x(40x^5 - 243x^3 - 7x^2 - 3x + 3)}{(40x^3 + x + 1)^3}$








Based from the graph of $f'$, we can estimate that $f'(x) > 0$ ($f$ is increasing) on the interval $(- \infty, -0.35) \bigcup (-0.2, \infty)$. Since $f(x)$ is always positive, we can say that the function don't have decreasing intervals.

Based from the graph of $f''$. We can estimate the intervals where the function has upward concavity, that is $f''(x) > 0$ on intervals $(- \infty, -0.5) \bigcup (0, 0.21)$

On the other hand, the intervals where the function has downward concavity are from $(0.05, 0) \bigcup (0.21, \infty)$. Therefore, the points of inflection can be approximated as $f(0) \approx - 1, f(0.21) \approx -0.70$.