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Let $x$ be a real number such that $x^2 + 7x + 12 \le 0.$ Find the largest possible value of $x^2 + 5x + 6.$

 

I gave up on this problem. The solution stated that once you factor the quadratic in the function (x+4)(x+3) and, you get -4<= x <= -3. Then you complete the square on x^2 + 5x + 6 to get \[x^2 + 5x + 6 = \left( x + \frac{5}{2} \right)^2 - \frac{1}{4}.\] Since the function decreased as it approaches -inf and increased as it approaches inf, the largest possible value of $x^2 + 5x + 6$ on the interval $-4 \le x \le -3$ occurs at $x = -4,$ which is $(-4)^2 + 5 \cdot (-4) + 6 = \boxed{2}.$

 

If the function increases when x increases, then why is the answer -4?

 Jun 6, 2024
 #1
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The first function will  be <  0  on [-4,  -3]

 

x^2 + 5x + 6  = ( x + 3) ( x + 2)

 

At  x = -3 ....   x^2 + 5x + 6  =  (-3 + 3) (-3 + 2)  =  0

 

At x = -4........x^2 + 5x + 6  =   (-4 + 3) (-4 + 2)  = 2

 

So  x = -4  is  correct

 

See the graph here : https://www.desmos.com/calculator/nf9utyhcye

 

 

cool cool cool

 Jun 6, 2024
edited by CPhill  Jun 6, 2024

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