+0  
 
-4
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avatar+318 

Find the critical values there should be 3:

 

f(x)= (3x^2 +3x +6 )/ (x^2+x-56)

 

help?????  i already got -1/2, and -7/75

 Jun 15, 2020
 #1
avatar+8342 
+2

\(f(x) = \dfrac{3x^2 + 3x + 6}{x^2 + x - 56} = 3 - \dfrac{58}5 \left(\dfrac{1}{x + 8} - \dfrac1{x - 7}\right)\)

 

Differentiating, \(f'(x) = \dfrac{58}5\left(\dfrac1{(x + 8)^2} - \dfrac1{(x - 7)^2}\right)\)

 

When \(f'(x) = 0\),

\((x - 7)^2 - (x + 8)^2 = 0\\ 2x + 1 = 0\\ x = -\dfrac12\)

 

Also, \(f\left(-\dfrac12\right) = -\dfrac7{75}\), so the critical point is \(\left(-\dfrac12, -\dfrac7{75}\right)\), and there is only 1 critical point, not 3.

 Jun 15, 2020
 #2
avatar+318 
0

When I enter -1/2 it says partially correct!! It is asking for another value!

mharrigan920  Jun 15, 2020
 #3
avatar+8342 
+1

The complete answer is (-1/2, -7/75), with the brackets. You don't just enter -1/2 and call it done.

MaxWong  Jun 15, 2020

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