# Is f(x)=5x^3-2x^2+5x+12 concave or convex at x=-1?

Mar 16, 2016

The function is concave at $f \left(- 1\right)$

#### Explanation:

A concave function is a function in which no line segment joining two points on its graph lies above the graph at any point.

A convex function, on the other hand, is a function in which no line segment joining two points on the graph lies below the graph at any point.

It means that, if $f \left(x\right)$ is more than the average of $f \left(x \pm \lambda\right)$ than the function is concave and if $f \left(x\right)$ is less than the average of $f \left(x \pm \lambda\right)$ than the function is convex.

Hence to find the convexity or concavity of $f \left(x\right) = 5 {x}^{3} - 2 {x}^{2} + 5 x + 12$ at $x = - 1$, let us evaluate $f \left(x\right)$ at $x = - 1.5 , - 1 \mathmr{and} - 0.5$.

$f \left(- 1.5\right) = 5 {\left(- \frac{3}{2}\right)}^{3} - 2 {\left(- \frac{3}{2}\right)}^{2} + 5 \left(- \frac{3}{2}\right) + 12 = - \frac{135}{8} - \frac{45}{4} - \frac{15}{2} + 12 = \frac{- 135 - 90 - 60 + 96}{8} = - \frac{189}{8}$

$f \left(- 1\right) = 5 {\left(- 1\right)}^{3} - 2 {\left(- 1\right)}^{2} + 5 \left(- 1\right) + 12 = - 5 - 2 - 5 + 12 = 0$

$f \left(- 0.5\right) = 5 {\left(- \frac{1}{2}\right)}^{3} - 2 {\left(- \frac{1}{2}\right)}^{2} + 5 \left(- \frac{1}{2}\right) + 12 = - \frac{5}{8} - \frac{2}{4} - \frac{5}{2} + 12 = \frac{- 5 - 4 - 20 + 96}{8} = \frac{67}{8}$

The average of $f \left(- 1.5\right)$ and $f \left(- 0.5\right)$ is $\frac{- \frac{189}{8} + \frac{67}{8}}{2} = - \frac{122}{2 \times 8} = - \frac{61}{8}$

As, this is less than $f \left(- 1\right)$, at $f \left(- 1\right)$ the function is concave.

graph{5x^3-2x^2+5x+12 [-2, 2, -20, 20]}