Chapter 4.3, Problem 47E

To determine

To estimate: The intervals of concavity to one decimal place by using a computer algebra system to compute and graph $f\text{'}\text{'}$ .

Answer to Problem 47E

The curve is concave downward on the intervals $\left(-\infty ,-0.6\right)$ and $\left(0,\infty \right)$ , and it is concave upward on the interval $\left(-0.6,0\right)$ .

Explanation of Solution

Given:

The function is $f\left(x\right)=\frac{x^4+x^3+1}{\sqrt{x^2+x+1}}$

Calculation:

Consider the function $f\left(x\right)=\frac{x^4+x^3+1}{\sqrt{x^2+x+1}}$ .

Use a computer algebra system, to compute and graph $f\text{'}\text{'}$ .

To compute the function $f\text{'}\text{'}$ , first define the function $f$ as follows.

  $f\left(x\right)=\frac{x^4+x^3+1}{\sqrt{x^2+x+1}}$

  $x\to \frac{x^4+x^3+1}{\sqrt{x^2+x+1}}$

Then, compute the value for $f\text{'}\text{'}\left(x\right)$ , by using the maple command $\text{diff}\left(\text{diff}\left(f\left(x,\right),x\right),x\right)$ . Then the output is as follows.

  $\frac{12x^2+6x}{\sqrt{x^2+x+1}}-\frac{\left(4x^3+3x^2\right)\left(2x+1\right)}{{\left(x^2+x+1\right)}^{3/2}}+\frac{3}{4}\frac{\left(x^4+x^3+1\right){\left(2x+1\right)}^2}{{\left(x^2+x+1\right)}^{5/2}}-\frac{x^4+x^3+1}{{\left(x^2+x+1\right)}^{3/2}}$

To plot the graph of the function $f\text{'}\text{'}$ , use the maple command as $\text{plot(diff(diff}\left(\text{f}\left(\text{x}\right)\text{,x}\right)\text{x}=-\mathrm{1.5..1.5},-\mathrm{3..10},\text{thickness}=3,\text{color}=\text{blue});$ .

  

From the graph of the function $f\text{'}\text{'}$ , it is observed that the curve is concave downward on the intervals $\left(-\infty ,-0.6\right)$ and $\left(0,\infty \right)$ , and it is concave upward on the interval $\left(-0.6,0\right)$ .