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Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. f(x) = x^3 - 20x^2 + 55 Enter the exact answers in increasing order. If there is no answer, enter "NA". The critical point at x = 0 is a local maximum The critical point at x = 16 is a local minimum The inflection point is x = 8

          Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither.
f(x) = x^3 - 20x^2 + 55
Enter the exact answers in increasing order. If there is no answer, enter "NA".
The critical point at x = 0 is a local maximum
The critical point at x = 16 is a local minimum
The inflection point is x = 8
        
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Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither.
f(x) = x^3 - 20x^2 + 55
Enter the exact answers in increasing order. If there is no answer, enter "NA".
The critical point at x = 0 is a local maximum
The critical point at x = 16 is a local minimum
The inflection point is x = 8

Added by Julie F.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, local minimum, or neither. x = 20x + 55 Enter the exact answers in increasing order. If there is no answer, enter "NA". The critical point at a local maximum. The critical point at x is a local minimum. The inflection point is x = .
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Transcript

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00:01 Hello students, to find the critical points of the function f of x is equal to x to the power 5 minus 10 x to the power 4 plus 30.
00:10 We will first find the first derivative that is f dash x then solve for x when f dash x is equal to 0.
00:16 So first f dash x is equal to d by dx of x to the power 5 minus 10 x to the power 4 plus 30 that is equal to 5 x to the power 4 minus 10 into 4 x to the power 3.
00:29 So 5 x to the power 4 minus 40 x cube.
00:34 So we have 5 x cube that is x minus 8.
00:40 So this gives us true critical points that is x is equal to 0 0 0 and 8 because x cube equal to 0 so 3 0s and x is equal to 8.
00:54 So we have 4 root that is 0 and 8.
00:59 So we can write critical points x is equal to 0 and 8.
01:10 Now moving forward to the second bit to find the inflection point we need to find where second derivative changes sign.
01:20 Let's find f double dash x.
01:21 So f double dash x is equal to 5 x to the power 4 minus 4 x cube.
01:32 So d by dx of 5 x to the power 4 minus 4 x cube.
01:37 So we have 20 x cube minus 120 x square that is equal to 20 x square equal to x minus 6.
01:53 So that gives us another critical point...
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