Answer:
R3 <= 0.083
Step-by-step explanation:
f(x)=xlnx,
The derivatives are as follows:
f'(x)=1+lnx,
f"(x)=1/x,
f"'(x)=-1/x²
f^(4)(x)=2/x³
Simialrly;
f(1) = 0,
f'(1) = 1,
f"(1) = 1,
f"'(1) = -1,
f^(4)(1) = 2
As such;
T1 = f(1) + f'(1)(x-1)
T1 = 0+1(x-1)
T1 = x - 1
T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2
T2 = 0+1(x-1)+1(x-1)^2
T2 = x-1+(x²-2x+1)/2
T2 = x²/2 - 1/2
T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3
T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3
T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)
Thus, T1(2) = 2 - 1
T1(2) = 1
T2 (2) = 2²/2 - 1/2
T2 (2) = 3/2
T2 (2) = 1.5
T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)
T3(2) = 4/3
T3(2) = 1.333
Since;
f(2) = 2 × ln(2)
f(2) = 2×0.693147 =
f(2) = 1.386294
Since;
f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).
Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,
Since;
f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2
Finally;
R3 <= |2/(4!)(2-1)^4|
R3 <= | 2 / 24× 1 |
R3 <= 1/12
R3 <= 0.083
Given :
C, D, and E are col-linear, CE = 15.8 centimetres, and DE= 3.5 centimetres.
To Find :
Two possible lengths for CD.
Solution :
Their are two cases :
1)
When D is in between C and E .
. . .
C D E
Here, CD = CE - DE
CD = 15.8 - 3.5 cm
CD = 12.3 cm
2)
When E is in between D and C.
. . .
D E C
Here, CD = CE + DE
CD = 15.8 + 3.5 cm
CD = 19.3 cm
Hence, this is the required solution.
Answer:
height=3.28m
Step-by-step explanation:
air means volume of room
let the height of room be x.
so, volume of room= 156.1 m³
8.31*5.72*x=156.1m³
x =3.28m
Answer:
2/3,1,3/2,9/4,27/8
Step-by-step explanation:
f(1)=2/3
f(n)=f(n-1)×3/2
f(2)=f(2-1)×3/2=f(1)×3/2=2/3×3/2=1
f(3)=f(3-1)×3/2=f(2)×3/2=1×3/2=3/2
f(4)=f(4-1)×3/2=f(3)×3/2=3/2×3/2=9/4
f(5)=f(5-1)×3/2=f(4)×3/2=9/4×3/2=27/8
