onegirl
Find all critical numbers and use the First Derivative Test to classify each as the location of a location maximum, local minimum, or neither. y = x^4 + 4x^3 - 2
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onegirl
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@electrokid can u help?
electrokid
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using our previous discussions, you can start the work and I guide.
electrokid
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step 1) first derivative
\(y'=?\)
onegirl
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Sorry i replied late i had internet issues. but the first derivative is 4x^2(x + 3)
electrokid
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good.
step 2) solve for "x":
\(4x^2(x+3)=0\)
onegirl
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okay i got x = -3 and x = 0
onegirl
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are you there?
electrokid
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good.
those are the critical points.
step 3) classify as maxima, minima or neither
step 3.1) find second derivative of "f" -> find f''(x)
onegirl
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how do i do that?
onegirl
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are u there @electrokid
electrokid
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step 3.1) find the second derivative.
onegirl
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okay
onegirl
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I got 12x^2 + 24x
PeterPan
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Remember the critical points that you got? If the second derivative is positive when evaluated at one of those critical points, that critical point would be a local minimum.
On the flip-side, if the second derivative is negative when evaluated at a critical point, that point would be a local maximum.
onegirl
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so i substitute those critical point into the second derivative
PeterPan
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Yes.
electrokid
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yes.
\(f''(x)=12x^2+24x\)
step 3.2) plug in x=-3 in this equation and check the sign..
onegirl
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okay
onegirl
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i got 36
onegirl
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and when i plugged in 0 i got 0
PeterPan
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Okay, so, f''(-3) = 36
This is positive, therefore...?
onegirl
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so it is a minimum?
PeterPan
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a local minimum ;)
What about f''(0) ?
onegirl
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I got 0
PeterPan
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It's neither positive nor negative, right?
onegirl
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yes
PeterPan
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Well, what does that mean, then?
onegirl
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so it will be neither
PeterPan
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It will be neither :D
onegirl
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Okay thanks
electrokid
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@PeterPan and @onegirl both did a great job and effort and both deserve a medal :)
so, @onegirl give one to P-Pan
☮
PeterPan
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Already done. ^.^
onegirl
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lol I already did