1) For this one I got the first box right, what would the second box answer be?
ID: 2831247 • Letter: 1
Question
1) For this one I got the first box right, what would the second box answer be?
2) I think it would be easy to not make the answer u provide less confusing by typing the whole sentence out and put the answer in the blanks buy putting (paranthesys) around the answer. Same apply with the rest of the problems.
3)
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Suppose that f(x) = x3 - 7x2 + 10. List the X values of all local maxima of f. If there are no local maxima, enter 'NONE' x values of local maximums = List the x values of all local minima of f. If there are no local minima, enter 'NONE'. x values of local minimums = The function f(x) = 9x + 3x-1 has one local minimum and one local maximum. This function has a local minimum at x equals with value and a local maximum at x equals with The function f(x) = 2x3 - 30x2 + 54x + 2 has one local minimum and one local maximum. This function has a local minimum at x equals with value and a local maximum at x equals with value The function f(x) = -2x3 + 30x2 - 126x + 1 has one local minimum and one local maximum. This function has a local minimum at x = with value f(x) = , and a local maximum at x = with value f(x) =Explanation / Answer
1) First let us compute f'(x).
f'(x)=3x^2-14x
Equating this to zero we get, x=0,x=14/3
Put this values in f''(x)=6x-14
local maxima=0....f''(x)<0
local minima x=14/3....f''(x)>0
2)Again we first find f'(x)
f'(x)=9-3/x^2
Equating this to 0,
x=sqrt(1/3) and
x=-sqrt(1/3)
f''(x)=6/(x^3).
local maxima=-sqrt(1/3)
local minima=sqrt(1/3).
3)Find f'(x),
f'(x)=6x^2-60x+54
Equating to 0.
x^2-10x+9=0
x=9, 1
f''(x)=2x-10
local maxima at x=1...f''(x)<0.
local minima x=9...f''(x)>0
4)Find f'(X)
6x^2+60x-126=0
x^2-10x+21=0
x=7,3
f''(x)=2x-10
local maxima at x=3
local minima at x=7
Hope this helps...plz rate....thanx...
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