Critical values

  • Thread starter apiwowar
  • Start date
  • #1
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f(x) = (1x+10)/(x^2+x+1)

so the question is to find the critical values for this function

i know that to find the critical values you have to take the derivative of the function. to do that you have to do the quotient rule

so you get

f'(x) = (x^2+x+1)(1)-(1x+10)(2x+1)/(x^2+x+1)^2

after cleaning it up i got (-x^2-20x-9)/(x^2+x+1)^2

so you set the numerator equal to 0 and solve it using the quadratic equation so after that you get

(20 +- sqrt (400 - 36))/-2

after simplifying i got

-10 +- sqrt364


but the website says that is wrong
where did i go wrong???
 

Answers and Replies

  • #2
35,004
6,757
f(x) = (1x+10)/(x^2+x+1)

so the question is to find the critical values for this function

i know that to find the critical values you have to take the derivative of the function. to do that you have to do the quotient rule

so you get

f'(x) = (x^2+x+1)(1)-(1x+10)(2x+1)/(x^2+x+1)^2

after cleaning it up i got (-x^2-20x-9)/(x^2+x+1)^2

so you set the numerator equal to 0 and solve it using the quadratic equation so after that you get

(20 +- sqrt (400 - 36))/-2

after simplifying i got

-10 +- sqrt364


but the website says that is wrong
where did i go wrong???

You forgot to divide sqrt(364) by 2. I get x = -10 +/- sqrt(91)
 
  • #3
LCKurtz
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Your last step should have (1/2)sqrt(364) and can be simplified.
 
  • #4
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how exactly did you get sqrt(91)?
364 divided by 2 is 182 so where does 91 come from?
 
  • #5
Office_Shredder
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because

[tex] \frac{\sqrt{364}}{2} = \frac{\sqrt{364}}{\sqrt{4}} = \sqrt{ \frac{364}{4}}[/tex]
 
  • #6
96
0
o ok, thanks alot
 

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