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

Question

Christopher Agresta · Accepted Answer

On the interval #[0,oo)#, we have absolute maxima at #x=2#, where #f(x)=1/5#

At absolute maximum or minimum derivative of #f(x)=(x+1)/(x^2+x+9)# will be zero. We have a maximum when second derivative is negative.

#(df)/(dx)=((x^2+x+9)xx1-(x+1)(2x+1))/(x^2+x+9)^2#

= #(x^2+x+9-2x^2-3x-1)/(x^2+x+9)^2#

= #-(x^2+2x-8)/(x^2+x+9)^2#

= #-((x+4)(x-2))/(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

#(d^2f)/(dx^2)=-((x^2+x+9)^2(2x+2)-2(x^2+x+9)(2x+1)(x^2+2x-8))/(x^2+x+9)^4#

= #-((x^2+x+9)((x^2+x+9)(2x+2)-(4x+2)(x^2+2x-8)))/(x^2+x+9)^4#

= #-(2x^3+4x^2+20x+18-(4x^3+10x^2-28x-16))/(x^2+x+9)^3#

= #(2x^3+6x^2-48x-34)/(x^2+x+9)^3#

at #x=2#, we have #(d^2f)/(dx^2)=-2/75#

and at #x=-4#, #(d^2f)/(dx^2)=2/147#

Hence we have an absolute maxima at #x=2# and #f(2)=3/15=1/5#

And as degree of denominator is higher, while at #x=0#, #f(0)=1/9#, as #x->oo#, #f(x)->0#

graph{x^2+x+9=0 [-10, 10, -0.4, 0.4]}

graph{x^2+x+9=0 [-200, 200, -0.4, 0.4]}

Ezra Adams · Answer

undefined To find the absolute maximum and minimum values of f(x) = (x + 1)/(x^2 + x + 9) on the interval [0, ∞), we need to follow these steps:

1. Find the critical points of f(x) by setting the derivative equal to zero and solving for x.
2. Evaluate f(x) at the critical points and endpoints of the interval.
3. Compare the values obtained in step 2 to determine the absolute maximum and minimum.

Let's proceed with these steps:

1. Differentiating f(x) with respect to x, we get:
f'(x) = [(x^2 + x + 9)(1) - (x + 1)(2x + 1)] / (x^2 + x + 9)^2
Simplifying, we have:
f'(x) = (x^2 + x + 9 - 2x^2 - 3x - 1) / (x^2 + x + 9)^2
f'(x) = (-x^2 - 2x + 8) / (x^2 + x + 9)^2

Setting f'(x) equal to zero, we solve for x:
-x^2 - 2x + 8 = 0
x^2 + 2x - 8 = 0
(x + 4)(x - 2) = 0
x = -4 or x = 2

2. Now, we evaluate f(x) at the critical points and endpoints of the interval:
f(0) = (0 + 1)/(0^2 + 0 + 9) = 1/9
f(2) = (2 + 1)/(2^2 + 2 + 9) = 3/15 = 1/5

3. Comparing the values obtained, we find:
f(0) = 1/9 is the absolute minimum value on the interval [0, ∞).
f(2) = 1/5 is the absolute maximum value on the interval [0, ∞).

Therefore, the absolute minimum value is 1/9 and the absolute maximum value is 1/5.