Answer
(a) Using the graph of $f(x)$, we can make the following estimates:
The graph is concave down in these intervals: $(-\infty, -1)\cup (-0.4, 0.6)$
The graph is concave up in these intervals: $(-1, -0.4)\cup (0.6, \infty)$
The points of inflection are $(-1, 0), (-0.4, 0.4),$ and $(0.6, 0.4)$
(b) Using the graph of $f''(x)$, we can make the following estimates:
The graph is concave down in these intervals: $(-\infty, -1)\cup (-0.29, 0.69)$
The graph is concave up in these intervals: $(-1, -0.29)\cup (0.69, \infty)$
The points of inflection are $(-1, 0), (-0.29, 0.60),$ and $(0.69, 0.46)$
Work Step by Step
(a) $f(x) = (x-1)^2(x+1)^3$
Using the graph of $f(x)$, we can make the following estimates:
The graph is concave down in these intervals: $(-\infty, -1)\cup (-0.4, 0.6)$
The graph is concave up in these intervals: $(-1, -0.4)\cup (0.6, \infty)$
The points of inflection are $(-1, 0), (-0.4, 0.4),$ and $(0.6, 0.4)$
(b) $f'(x) = 2(x-1)(x+1)^3+3(x-1)^2(x+1)^2$
$f''(x) = 2(x+1)^3+12(x-1)(x+1)^2+6(x-1)^2(x+1)$
Using the graph of $f''(x)$, we can make the following estimates:
The graph is concave down in these intervals: $(-\infty, -1)\cup (-0.29, 0.69)$
The graph is concave up in these intervals: $(-1, -0.29)\cup (0.69, \infty)$
We can find the points of inflection:
$f(-1) = (-1-1)^2(-1+1)^3 = 0$
$f(-0.29) = (-0.29-1)^2(-0.29+1)^3 = 0.60$
$f(0.69) = (0.69-1)^2(0.69+1)^3 = 0.46$
The points of inflection are $(-1, 0), (-0.29, 0.60),$ and $(0.69, 0.46)$
