Answer:-24
explain how you know:
Since xy = 4, we have y=4/x. Substituting in (1) we get f(x) = 4x+(36/x) …….(2)
Differentiating we get f ' (x) = 4 - (36/x^2)…..(2)
For maxima or minima, f ‘ (x) = 0
=> 4 - (36/x^2) =0 => (4x^2 - 36)/x^2 = 0 which gives x= +/- 3
Differentiating (2), f ‘’ (x) = 72/x^3
When x= +3 , f ‘’ (x) is clearly +ve. Therefore f(x) at (2) gives minima and the minimum value is 24.
When x= -3, f ‘’ (x) is -ve. Therefore, f(x) has maxima at x= -3. The maximum value is -24.
Note: You may be surprised to observe that the minima is more than the maxima. This is due to the fact that the function is discontinuous at x= 0, the graph of f(x) comprising of two branches, one in the first quadrant giving the minima (= +24) and another in the third quadrant that gives maxima (= -24)
Answer:
a) aₙ = 4.(2)^n-1
b) aₙ = 2+3(n-1)
c) aₙ = 3+4(n-1)
d) aₙ = 3.(4)^n-1
Step-by-step explanation:
a) a geometric sequence with first term of 4 and a common ratio of 2
aₙ = ar^n-1
where a is the first term and r is the common ratio
aₙ = 4.(2)^n-1
Comparing with the standard form:
1st term = 4
Common ratio = 2
b) an arithmetic sequence with a first term of 2 and a common difference of 3
The standard form of arithmetic sequence is:
aₙ = a₁ + d(n-1)
where a₁ is the first term and d is the common difference
Given: aₙ = 2+3(n-1)
Comparing with standard equation
First term = 2
Common difference = 3
c) an arithmetic sequence with a first term of 3 and a common difference of 4
The standard form of arithmetic sequence is:
aₙ = a₁ + d(n-1)
where a₁ is the first term and d is the common difference
Given: aₙ = 3+4(n-1)
Comparing with standard equation
First term = 2
Common difference = 3
d) a geometric sequence with first term of 3 and a common ratio of 4
aₙ = ar^n-1
where a is the first term and r is the common ratio
aₙ = 3.(4)^n-1
Comparing with the standard form:
1st term = 3
Common ratio = 4
I dont know exactly but its like 160÷2 - 11
X-3y, x=3, y=-2
(3)-3(-2)
(3)+(3*2)
3+6=9
