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missing critical points with some equations?(most/all work done)?
Iv been studying calculus, and I understand pretty well how to graph a function, but in soem problems thre are these seemingly hidden critical points
Example: f(x)= x^4-2x^3
f'(x)=4x^3-6x^2=0
4x^3/4x^2=6x^2/4x^2
x=3/2
f''(x)=12x^2-12x
concativity at 3/2 is +
12x^2-12x=0
12x^2=12x
x=1(conacativity is - when less, + when more)
Punch into f(x) to get points blah blah....so 2 and 0 are also critical points, which inhernently makes sense looking at f(x), but there has to be something with my method. I have my final tommorow, and this is rly gonna worry me. plz help
sorry, not critical. inflection points. well thats what it looks like to me, my answer key is a set of microscopic graphs
not so much at 2, but x=0 is definately shaped like an inflection point. I think I mite get it if 2 is just an intercept that theyre graphing in the answer key. So i guess teh only thing you have to answer is this: is the only way to tell if x=0 is a critical/inflection point by looking at it? easy points...
1 Answer
- ?Lv 77 years agoFavorite Answer
==> critical number: call it c
f '(c) = 0 or f '(c) = undefined
c must be within the domain of f(x)
f(x) = x⁴ - 2x³
f '(x) = 4x³ - 6x²
f '(c) = 4c³ - 6c² = 0
4c³ - 6c² = 0
2c²(2c - 3) = 0
2c² = 0, 2c - 3 = 0
c = 0 and c = 3/2
f(x) = x⁴ - 2x³
f(c) = c⁴ - 2c³
f(0) = 0 and f(3) = 81/16 - 108/16 = -27/2
(0,0) and (3,27)
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==> inflection points: where x = a
f ''(a) = 0
f(x) = x⁴ - 2x³
f '(x) = 4x³ - 6x²
f ''(x) = 12x² - 12x
12c² - 12c = 0
12c(c - 1) = 0
c = 0 and c = 1
f(x) = x⁴ - 2x³
f(a) = a⁴ - 2a³
f(0) = 0⁴ - 2*0³ = 0
f(1) = 1⁴ - 2*1³ = -1
(0,0) and (1, -1)
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==> concavity
take the x-values of the inflection points and test the points around them after determining the intervals using x-values of the inflection points;
a = 0 and a = 1
intervals on the number-line
-∞ .............. 0 ......1.......... ∞
(-∞, 0) , (0, 1) and (1, ∞)
(-∞, 0) take any value between -∞ and 0; x = -1, f ''(x) = 12x² - 12x = 12(-1)² - 12(-1) = 24 ==> concave up in this interval
(0, 1) take any value between 0 and 1; x = ½, f ''(x) = 12x² - 12x = 12(½)² - 12(½) = -3 ==> concave down in this interval
(1,∞) take any value between 0 and 1; x = 10, f ''(x) = 12x² - 12x = 12(10)² - 12(10) = 1080 ==> concave up in this interval