CPM Homework Help : APCALC Problem 8-130

Multiple Choice: Given $f(x)=\frac { 1 } { x }- 4x^2 + 7x$ , then $f$ is concave up for:

$x ≠ 0$

no values of $x$

$x < 0$

$x < 0 \text { or } x > \frac { 1 } { \sqrt [ 3 ] { 4 } }$

$0 < x < \frac { 1 } { \sqrt [ 3 ] { 4 } }$

Hint:

Recall that candidates for points of inflection occur where the second derivative equals zero AND where the second derivative does not exist.

Steps:

After you identify where $f^{\prime\prime}(x) = 0$ and where $f^{\prime\prime}(x) = DNE$ , check the regions before, after and between those candidates: if $f^{\prime\prime}(x) > 0$ , then $f(x)$ is concave up.