Asked by mkcolema
The function 9x+3x^-1 has one local minimum x=?? and one local maximum x=??.
Answers
Answered by
GanonTEK
f[x] = 9x + 3/x
f'[x] = 9 - 3/(x^2)
f'[x] = 0 at min/max
9 - 3/(x^2) = 0
multiply across by x^2 to get
9x^2 - 3 = 0
x^2 = 1/3
x = 1/sqrt(3) or x = -1/sqrt(3)
which i will call x1 and x2.
to find which is the min and which is the max get f''[x] and sub in the two above values for x
So if f''[x1]>0 that point is a min
if f''[x1]<0 that point is a max
repeat for x2.
to find the y coordinates sub x1 and x2 back into f[x].
ok hope that helps
f'[x] = 9 - 3/(x^2)
f'[x] = 0 at min/max
9 - 3/(x^2) = 0
multiply across by x^2 to get
9x^2 - 3 = 0
x^2 = 1/3
x = 1/sqrt(3) or x = -1/sqrt(3)
which i will call x1 and x2.
to find which is the min and which is the max get f''[x] and sub in the two above values for x
So if f''[x1]>0 that point is a min
if f''[x1]<0 that point is a max
repeat for x2.
to find the y coordinates sub x1 and x2 back into f[x].
ok hope that helps
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