Determine the location of each local extremum of the function
$f(x)= -x^3 + 12x^2 - 45x - 2$
Critical values I found are $x=3$, and $x=5$
Critical points I found are $(3, -56) (5, -52)$
How do I find the local minimum/maximum?
How do I find the local extrema with the first derivative test?
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calculus
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0You found them: $f(3)=-56$ is a local minimum, and $f(5)=-52$ is a local maximum. – 2017-01-31
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0That's weird, I entered my answer but they seem to be wrong – 2017-01-31
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0I don't know what form(s) you are working with, but doublecheck what is being asked. Your terminology is a bit unusual, I'd have called $3, 5$ critical points, and $-56,-52$ critical values (or local extrema). – 2017-01-31
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0I don't know what i'm doing really but trying to follow examples from the class – 2017-01-31
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0It's asking for the local minimum and to type an ordered pair, I put (3, -56) but don't know why its wrong – 2017-01-31
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0Ok, I'm totally confused now, the correct answer is supposed to be (-5, 48) how do they get that? – 2017-01-31
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0Your numbers are correct for the posted equation, as can be verified [here](http://www.wolframalpha.com/input/?i=-x%5E3+%2B+12x%5E2+-+45x+-+2) (towards the bottom of the page). – 2017-01-31
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0Thanks a lot I figured what went wrong – 2017-01-31
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0Let me guess: $(-5,48)$ is the right answer for $f(x) = -x^3 + 12x^2 - 45x + 2$ ;-) – 2017-01-31
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1=P its f(x)= −x^3-12x^2−45x−2 – 2017-01-31
2 Answers
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Well because theres only two local extremums, the local maximum must be at $(5, -52)$ and the local minimum is at $(3, -56)$, as the cubic function is unbounded at the ends of (-$\infty$, $\infty$).
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The local minimum or local maximum are points on your f(x) graph where there's a valley. You can find local maximum and local minimum value by using your critical points and plugging them into the second derivative. If the number f''(x)<0, then (x,f(x)) is a local max and if f''(x)>0, then it's a local min.