#Finding Absolute Extrema

15 messages · Page 1 of 1 (latest)

serene thistle
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I need some help on a question to find the Abs max and min. I normally just put the end points and then the hills/valleys into the equation but for this one it isn't working out. Can someone help solve this please?

hollow vortexBOT
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serene thistle
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Let me send a picture

dawn aspen
# serene thistle

$f(0)=\boxed{0}$, $f(4)=\frac{64}{3}-40+24=\frac{64}{3}-16=\boxed{\frac{16}{3}}$.\
$f'(x)=x^2-5x^2+6=(x-2)(x-3)$, so critical points are at $x=2, 3$\
$f(2)=\frac83-10+12=\frac83+2=\boxed{\frac{14}{3}}$\
$f(3)=9-\frac{45}{2}+18=27-\frac{45}{2}=\boxed{\frac92}$\
since $f(0)<f(3)<f(2)<f(4)$ we have that $\boxed{\text{the absolute maxima occurs at}\hspace{1mm} x=4 \hspace{1mm}\text{at the coordinate}\hspace{1mm} \left(4, \frac{16}{3}\right),\hspace{1mm} \text{and the absolute minima occurs at}\hspace{1mm} x=0\hspace{1mm} \text{at the coordinate}\hspace{1mm} (0, 0)}$.

serene thistle
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I'm looking at my notes and I don't understand what these critical points are, like what's the purpose of them

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Like you have end points and you have points that are the "hills and valleys" in between the end points.

unreal zodiacBOT
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vengeance

1350294910867210291.png
dawn aspen
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so either local maximum, minimum, or inflection point occurs there

serene thistle
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Oh so we get the derivative of the equation, set it equal to zero and simply factor?

dawn aspen
serene thistle
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Alright that makes sense

deft meteorBOT
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@serene thistle

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