Question

If $y=x^4+x^3-18x^2+24x-12$ then

• A. $(-2,24)$ is a point of inflection
• B. $(-2,3/2)$ is a point of inflection
• C. $(3/2,-8\frac{1}{16})$ is a point of inflection
• D. $y$ is convex on $(3/2,\infty )$

Answer 1

The correct option is C $(3/2,-8\frac{1}{16})$ is a point of inflection

$y=x^4+x^3-18x^2+24x-12$
$y^{\prime }=4x^3+3x^2-36x+24$
$y^{\prime \prime }=12x^2+6x-36=12(x^2+\frac{x}{2}-3)$
hence $y^{\prime \prime }=0$ at $x_1=-2$, $x_2=3/2$
Also $y">0$ on $(-\infty ,-2)$ and $(3/2,\infty ,)$
and $y"<0$ on $(-2,3/2)$
so the points $(\frac{3}{2},-8\frac{1}{16})$ are points of inflection and $y$ is concave on $(3/2,\infty )$.
Ans: C