# How do you determine whether the function f (x) = x sqrt(x^2+2x+5)+1 sqrt(x^2+2x+5) is concave up or concave down and its intervals?

Jul 30, 2015

It's concave up for $x > - 1$ (on the interval $\left(- 1 , \infty\right)$) and concave down for $x < - 1$ (on the interval $\left(- \infty , - 1\right)$).

#### Explanation:

Write the function as $f \left(x\right) = \left(x + 1\right) \sqrt{{x}^{2} + 2 x + 5}$ and then use the Product Rule and the Chain Rule to get the first derivative:

$f ' \left(x\right) = \sqrt{{x}^{2} + 2 x + 5} + \left(x + 1\right) \cdot \frac{1}{2} \cdot {\left({x}^{2} + 2 x + 5\right)}^{- \frac{1}{2}} \cdot \left(2 x + 2\right)$

$= \frac{{x}^{2} + 2 x + 5 + {x}^{2} + 2 x + 1}{\sqrt{{x}^{2} + 2 x + 5}} = \frac{2 {x}^{2} + 4 x + 6}{\sqrt{{x}^{2} + 2 x + 5}}$

Now use the Quotient Rule and Chain Rule to find the second derivative:

$f ' ' \left(x\right)$

$= \frac{\sqrt{{x}^{2} + 2 x + 5} \cdot \left(4 x + 4\right) - \left(2 {x}^{2} + 4 x + 6\right) \cdot \frac{1}{2} \cdot {\left({x}^{2} + 2 x + 5\right)}^{- \frac{1}{2}} \cdot \left(2 x + 2\right)}{{x}^{2} + 2 x + 5}$

$= \frac{\left({x}^{2} + 2 x + 5\right) \left(4 x + 4\right) - \left(2 {x}^{2} + 4 x + 6\right) \left(x + 1\right)}{{\left({x}^{2} + 2 x + 5\right)}^{\frac{3}{2}}}$

$= \frac{4 {x}^{3} + 8 {x}^{2} + 20 x + 4 {x}^{2} + 8 x + 20 - 2 {x}^{3} - 4 {x}^{2} - 6 x - 2 {x}^{2} - 4 x - 6}{{\left({x}^{2} + 2 x + 5\right)}^{\frac{3}{2}}}$

So

$f ' ' \left(x\right) = \frac{2 {x}^{3} + 6 {x}^{2} + 18 x + 14}{{\left({x}^{2} + 2 x + 5\right)}^{\frac{3}{2}}}$

Since ${x}^{2} + 2 x + 5 > 0$ for all real numbers $x$, the sign (positive or negative) of $f ' ' \left(x\right)$ is determined by the sign of $2 {x}^{3} + 6 {x}^{2} + 18 x + 14$, which is determined by the sign of ${x}^{3} + 3 {x}^{2} + 9 x + 7$.

The number $x = - 1$ is a real root of ${x}^{3} + 3 {x}^{2} + 9 x + 7$ since $- 1 + 3 - 9 + 7 = 0$. If you use long-division of polynomials (or synthetic division), you'll find that ${x}^{3} + 3 {x}^{2} + 9 x + 7 = \left(x + 1\right) \left({x}^{2} + 2 x + 7\right)$. The quadratic ${x}^{2} + 2 x + 7$ is always positive for all real $x$.

All of this implies that the sign of $f ' ' \left(x\right)$ is determined by the sign of $x + 1$. Since $x + 1 > 0$ when $x > - 1$ and $x + 1 < 0$ when $x < - 1$, the same is true for $f ' ' \left(x\right)$.

This means the graph of $f \left(x\right)$ is concave up for $x > - 1$ (on the interval $\left(- 1 , \infty\right)$) and concave down for $x < - 1$ (on the interval $\left(- \infty , - 1\right)$).