Q. 45 - Calculus

Question

Use a sign chart for $f\text{'}\text{'}$ to determine the intervals on which each function f in Exercises 41–52 is concave up or concave down, and identify the locations of any inflection points. Then verify your algebraic answers with graphs from a calculator or graphing utility 

$f\left(x\right)=\frac{1}{x^2+x+1}$

Step-by-Step Solution

Answer

Inflection is at $x=0\text{ and }x=-1$ and concave up on $(-\infty ,-1),(0,\infty )\text{ and concave down on }(-1,0).$

Step 1. Given Information.

The given function is $f\left(x\right)=\frac{1}{x^2+x+1}$.

Step 2. Second Derivative.

On differentiating the function, we get,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\frac{1}{x^2+x+1} \\ =\frac{-(2x+1)}{{\left(x^2+x+1\right)}^2} \\ {f}^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\frac{-(2x+1)}{{\left(x^2+x+1\right)}^2} \\ =\frac{\left(-\left(2\frac{d}{dx}x+\frac{d}{dx}1\right)\right){\left(x^2+x+1\right)}^2+(2x+1)\left(2\left(x^2+x+1\right)\frac{d}{dx}\left(x^2+x+1\right)\right)}{{\left(x^2+x+1\right)}^4} \\ =\frac{-2{\left(x^2+x+1\right)}^2+(2x+1)\left(2\left(x^2+x+1\right)(2x+1)\right)}{{\left(x^2+x+1\right)}^4} \\ =\frac{6x(x+1)}{{\left(x^2+x+1\right)}^3} \end{gathered}$$

Step 3. Sign Chart.

The second derivative is zero at $x=0,-1$

Therefore, the sign chart will be,

Therefore, the inflection points are $x=0,-1.$

Concave up on $(-\infty ,-1),(0,\infty )\text{ and concave down on }(-1,0).$

Step 4. Verification.

The graph of the function is ,