Critical Number Problem - # 4

J

Jason76

Senior Member
Joined
Oct 19, 2012
Messages
1,180
\(\displaystyle f(x) = \dfrac{x}{x^{2} - x + 16} \) Evaluated at interval \(\displaystyle [0, 12]\)

\(\displaystyle f'(x) = \dfrac{(x^{2} - x + 16)(1) - (x)(2x - 1)}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{x^{2} - x + 16 - 2x^{2} - 2}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{x^{2} - x + 16 - 2x^{2} - 2}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{-x^{2} - x + 14}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle \dfrac{-x^{2} - x + 14}{(x^{2} - x + 16)^{2}} = 0 \)

\(\displaystyle (x^{2} - x + 16)^{2} = 0 \)

\(\displaystyle x = -4.5\)

\(\displaystyle x = 3.5\)
 
Last edited:
Jason76 said:
\(\displaystyle f(x) = \dfrac{x}{x^{2} - x + 16} \)

\(\displaystyle f'(x) = \dfrac{(x^{2} - x + 16)(1) - (x)(2x - 1)}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{x^{2} - x + 16 - 2x^{2} - 2}{(x^{2} - x + 16)^{2}} \) ................................Incorrect

\(\displaystyle f'(x) = \dfrac{x^{2} - x + 16 - 2x^{2} - 2}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{-x^{2} - x + 14}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle \dfrac{-x^{2} - x + 14}{(x^{2} - x + 16)^{2}} = 0 \)

\(\displaystyle (x^{2} - x + 16)^{2} = 0 \)
Click to expand...
.
 
:cool: corrected

\(\displaystyle f(x) = \dfrac{x}{x^{2} - x + 16} \)

\(\displaystyle f'(x) = \dfrac{[(x^{2} - x + 16)(1)] - [(x)(2x - 1)}{(x^{2} - x + 16)^{2}]} \)

\(\displaystyle f'(x) = \dfrac{[x^{2} - x + 16] - [2x^{2} - 2]}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{x^{2} - x + 16 - 2x^{2} + 2}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{-x^{2} - x + 18}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle \dfrac{-x^{2} - x + 18}{(x^{2} - x + 16)^{2}} = 0 \)

\(\displaystyle (-x^{2} - x + 18) = 0 \)

\(\displaystyle x = -4.7\)

\(\displaystyle x = 3.7\) :confused:
 
Last edited:
Jason76 said:
:cool: corrected

\(\displaystyle f(x) = \dfrac{x}{x^{2} - x + 16} \)

\(\displaystyle f'(x) = \dfrac{[(x^{2} - x + 16)(1)] - [(x)(2x - 1)}{(x^{2} - x + 16)^{2}]} \)

\(\displaystyle f'(x) = \dfrac{[x^{2} - x + 16] - [2x^{2} - 2]}{(x^{2} - x + 16)^{2}} \) ...................................Incorrect

\(\displaystyle f'(x) = \dfrac{x^{2} - x + 16 - 2x^{2} + 2}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{-x^{2} - x + 18}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle \dfrac{-x^{2} - x + 18}{(x^{2} - x + 16)^{2}} = 0 \)

\(\displaystyle (-x^{2} - x + 18) = 0 \)

\(\displaystyle x = -4.7\)

\(\displaystyle x = 3.7\) :confused:
Click to expand...
.
 
:cool: corrected

\(\displaystyle f(x) = \dfrac{x}{x^{2} - x + 16} \)

\(\displaystyle f'(x) = \dfrac{[(x^{2} - x + 16)(1)] - [(x)(2x - 1)}{(x^{2} - x + 16)^{2}]} \)

\(\displaystyle f'(x) = \dfrac{[x^{2} - x + 16] - [2x^{2} - 1]}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{x^{2} - x + 16 - 2x^{2} + 1}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{-x^{2} - x + 17}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle \dfrac{-x^{2} - x + 17}{(x^{2} - x + 16)^{2}} = 0 \)

\(\displaystyle x = 3.6\)

\(\displaystyle x = -4.6\) :confused: These two values wrong on computer.
 
Jason76 said:
:cool: corrected

\(\displaystyle f(x) = \dfrac{x}{x^{2} - x + 16} \)

\(\displaystyle f'(x) = \dfrac{[(x^{2} - x + 16)(1)] - [(x)(2x - 1)}{(x^{2} - x + 16)^{2}]} \)

\(\displaystyle f'(x) = \dfrac{[x^{2} - x + 16] - [2x^{2} - 1]}{(x^{2} - x + 16)^{2}} \) .................Incorrect

\(\displaystyle f'(x) = \dfrac{x^{2} - x + 16 - 2x^{2} + 1}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{-x^{2} - x + 17}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle \dfrac{-x^{2} - x + 17}{(x^{2} - x + 16)^{2}} = 0 \)

\(\displaystyle x = 3.6\)

\(\displaystyle x = -4.6\)
Click to expand...
.
 
:cool: corrected

\(\displaystyle f(x) = \dfrac{x}{x^{2} - x + 16} \)

\(\displaystyle f'(x) = \dfrac{[(x^{2} - x + 16)(1)] - [(x)(2x - 1)}{(x^{2} - x + 16)^{2}]} \)

\(\displaystyle f'(x) = \dfrac{[x^{2} - x + 16] - [2x^{2} - x]}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{x^{2} - x + 16 - 2x^{2} + x}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{-x^{2} + 16}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle \dfrac{-x^{2} + 16}{(x^{2} - x + 16)^{2}} = 0 \)

\(\displaystyle (-x^{2} + 16) = 0 \)

\(\displaystyle -x^{2} = -16\)

\(\displaystyle x = \pm 4\) :confused: still wrong on computer
 
\(\displaystyle f(x) = \dfrac{x}{x^{2} - x + 16} \)

\(\displaystyle f'(x) = \dfrac{[(x^{2} - x + 16)(1)] - [(x)(2x - 1)}{(x^{2} - x + 16)^{2}]} \)

\(\displaystyle f'(x) = \dfrac{[x^{2} - x + 16] - [2x^{2} - x]}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{x^{2} - x + 16 - 2x^{2} + x}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{-x^{2} + 16}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle \dfrac{-x^{2} + 16}{(x^{2} - x + 16)^{2}} = 0 \)

\(\displaystyle (-x^{2} + 16) = 0 \)

\(\displaystyle -x^{2} = -16\)

\(\displaystyle x = \pm 4\) :confused: R these the critical numbers

Testing critical numbers (1st derivative test)

For below \(\displaystyle -4\)

\(\displaystyle f'(x) = \dfrac{-(-5)^{2} + 16}{((-5)^{2} - (-5) + 16)^{2}} = + \)


above \(\displaystyle -4\) below \(\displaystyle 4\)

\(\displaystyle f'(x) = \dfrac{-(-3)^{2} + 16}{((-3)^{2} - (-3) + 16)^{2}} = + \)

For above \(\displaystyle 4\)

\(\displaystyle f'(x) = \dfrac{-(5)^{2} + 16}{((5)^{2} - (5) + 16)^{2}} = - \)
 
For \(\displaystyle 0\)

\(\displaystyle f(0) = \dfrac{(0)}{(0)^{2} - (0) + 16} = 0 \)

For \(\displaystyle 4\)

\(\displaystyle f(4) = \dfrac{(4)}{(4)^{2} - (4) + 16} = \dfrac{4}{28} = \dfrac{1}{7} = .1429\)

For \(\displaystyle 12\)

\(\displaystyle f(12) = \dfrac{(12)}{(12)^{2} - (12) + 16} = \dfrac{12}{148} = \dfrac{3}{37} = .0811 \) - Is this the max?:confused:
 
Last edited:
Jason76 said:
\(\displaystyle f(x) = \dfrac{x}{x^{2} - x + 16} \)

\(\displaystyle f'(x) = \dfrac{[(x^{2} - x + 16)(1)] - [(x)(2x - 1)}{(x^{2} - x + 16)^{2}]} \)

\(\displaystyle f'(x) = \dfrac{[x^{2} - x + 16] - [2x^{2} - x]}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{x^{2} - x + 16 - 2x^{2} + x}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle f'(x) = \dfrac{-x^{2} + 16}{(x^{2} - x + 16)^{2}} \)

\(\displaystyle \dfrac{-x^{2} + 16}{(x^{2} - x + 16)^{2}} = 0 \)

\(\displaystyle (-x^{2} + 16) = 0 \)

\(\displaystyle -x^{2} = -16\)

\(\displaystyle x = \pm 4\) :confused: Are these the critical numbers? Yes, for the unrestricted function.

Testing critical numbers (1st derivative test)

For below \(\displaystyle -4\)

\(\displaystyle f'(x) = \dfrac{-(-5)^{2} + 16}{((-5)^{2} - (-5) + 16)^{2}} = + \ \ \ \)No. \(\displaystyle \ \ -(-5)^2 + 16 = -(25) + 16 = -\)


above \(\displaystyle -4\) below \(\displaystyle 4\)

\(\displaystyle f'(x) = \dfrac{-(-3)^{2} + 16}{((-3)^{2} - (-3) + 16)^{2}} = + \)

For above \(\displaystyle 4\)

\(\displaystyle f'(x) = \dfrac{-(5)^{2} + 16}{((5)^{2} - (5) + 16)^{2}} = - \)
Click to expand...
Jason76 said:
For \(\displaystyle 0\)

\(\displaystyle f(0) = \dfrac{(0)}{(0)^{2} - (0) + 16} = 0 \)

For \(\displaystyle 4\)

\(\displaystyle f(4) = \dfrac{(4)}{(4)^{2} - (4) + 16} = \dfrac{4}{28} = \dfrac{1}{7} = .1429\)

For \(\displaystyle 12\)

\(\displaystyle f(12) = \dfrac{(12)}{(12)^{2} - (12) + 16} = \dfrac{12}{148} = \dfrac{3}{37} = .0811 \) - Is this the max?:confused:\(\displaystyle \ \ \ \ \)Jason76, *why* would you ask that when f(4) above is larger!?
Click to expand...
.
 
Last edited:
You must log in or register to reply here.
Top