### Home > APCALC > Chapter 8 > Lesson 8.3.3 > Problem8-130

8-130.

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

1. $x ≠ 0$

1. no values of $x$

1. $x < 0$

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

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

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

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.