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
You can soustracte the first or the second one
and y cancel
x = 6
now you put 6 in x
you can take the first or the second one
6 - y = -2
6 - y = -2
y = -8
Answer:
The solutions of the equation are 0 and 0.75.
Step-by-step explanation:
Given : Equation 
To find : All solutions of the equation algebraically. Use a graphing utility to verify the solutions graphically ?
Solution :
Equation
Either 
or 
When
When
Solve by quadratic formula, 
The solutions of the equation are 0 and 0.75.
For verification,
In the graph where the curve cut x-axis is the solution of the equation.
Refer the attached figure below.
Answer:
x = 2
Step-by-step explanation:
Rewrite the equation as 4x - 6 = 2
Move all terms not containing x to the right side of the equation:
Add 6 to both sides of the equation.
4x = 2 + 6
Add 2 and 6.
4x = 8
Divide each term by 4 and simplify.
Divide each term in 4x = 8 by 4.
Reduce the expression by cancelling the common factors.
Divide 8 by 4.
x = 2
