# Find the absolute maximum and the absolute minimum values of f(x) = (x + 1)/(x^2 + x + 9) on the interval [ 0,∞)?

Apr 24, 2017

On the interval $\left[0 , \infty\right)$, we have absolute maxima at $x = 2$, where $f \left(x\right) = \frac{1}{5}$

#### Explanation:

At absolute maximum or minimum derivative of $f \left(x\right) = \frac{x + 1}{{x}^{2} + x + 9}$ will be zero. We have a maximum when second derivative is negative.

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\left({x}^{2} + x + 9\right) \times 1 - \left(x + 1\right) \left(2 x + 1\right)}{{x}^{2} + x + 9} ^ 2$

= $\frac{{x}^{2} + x + 9 - 2 {x}^{2} - 3 x - 1}{{x}^{2} + x + 9} ^ 2$

= $- \frac{{x}^{2} + 2 x - 8}{{x}^{2} + x + 9} ^ 2$

= $- \frac{\left(x + 4\right) \left(x - 2\right)}{{x}^{2} + x + 9} ^ 2$

as such we have extrema at $x = - 4$ and $x = 2$ and in the interval we have just one extrema at $x = 2$ and

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = - \frac{{\left({x}^{2} + x + 9\right)}^{2} \left(2 x + 2\right) - 2 \left({x}^{2} + x + 9\right) \left(2 x + 1\right) \left({x}^{2} + 2 x - 8\right)}{{x}^{2} + x + 9} ^ 4$

= $- \frac{\left({x}^{2} + x + 9\right) \left(\left({x}^{2} + x + 9\right) \left(2 x + 2\right) - \left(4 x + 2\right) \left({x}^{2} + 2 x - 8\right)\right)}{{x}^{2} + x + 9} ^ 4$

= $- \frac{2 {x}^{3} + 4 {x}^{2} + 20 x + 18 - \left(4 {x}^{3} + 10 {x}^{2} - 28 x - 16\right)}{{x}^{2} + x + 9} ^ 3$

= $\frac{2 {x}^{3} + 6 {x}^{2} - 48 x - 34}{{x}^{2} + x + 9} ^ 3$

at $x = 2$, we have $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = - \frac{2}{75}$

and at $x = - 4$, $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \frac{2}{147}$

Hence we have an absolute maxima at $x = 2$ and $f \left(2\right) = \frac{3}{15} = \frac{1}{5}$

And as degree of denominator is higher, while at $x = 0$, $f \left(0\right) = \frac{1}{9}$, as $x \to \infty$, $f \left(x\right) \to 0$

graph{y-(x+1)/(x^2+x+9)=0 [-10, 10, -0.4, 0.4]}

graph{y-(x+1)/(x^2+x+9)=0 [-200, 200, -0.4, 0.4]}